CLASS 11 PHYSICS CHAPTER 13 – OSCILLATIONS
NCERT CLASS 11 PHYSICS
CHAPTER 13 – OSCILLATIONS
IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS
1. What is periodic motion?
Answer:
Periodic motion is a type of motion in which an object repeats its motion after equal intervals of time.
The time taken to complete one cycle is called the time period.
Examples:
Motion of the Earth around the Sun.
Motion of the hands of a clock.
Oscillation of a simple pendulum.
2. What is oscillatory motion?
Answer:
Oscillatory motion is the repeated to-and-fro motion of an object about its mean or equilibrium position.
Examples:
Motion of a simple pendulum.
Motion of a mass attached to a spring.
Vibrations of a tuning fork.
3. Differentiate between periodic motion and oscillatory motion.
Answer:
Periodic motion:
Motion that repeats itself after equal intervals of time.
It may be circular, rotational or oscillatory.
Example: Revolution of the Earth around the Sun.
Oscillatory motion:
Motion in which an object moves repeatedly to and fro about a mean position.
It is a special type of periodic motion when the oscillations repeat regularly.
Example: Motion of a simple pendulum.
4. What is simple harmonic motion (SHM)?
Answer:
Simple harmonic motion is a special type of oscillatory motion in which the restoring force or acceleration is directly proportional to the displacement from the mean position and is always directed towards the mean position.
Mathematically,
F = −kx
or
a = −ω²x
where:
F = Restoring force
k = Force constant
x = Displacement
a = Acceleration
ω = Angular frequency
5. What is the mean position in oscillatory motion?
Answer:
The mean position is the equilibrium position of an oscillating body where the net restoring force is zero.
At this position:
Displacement is zero.
Speed is maximum in ideal SHM.
Acceleration is zero.
6. What is amplitude?
Answer:
Amplitude is the maximum displacement of an oscillating particle from its mean position.
It is represented by A.
SI unit: metre (m).
7. Define time period and frequency.
Answer:
Time Period (T):
The time taken by an oscillating particle to complete one full oscillation is called its time period.
SI unit: second (s).
Frequency (ν):
The number of complete oscillations made by a particle in one second is called frequency.
SI unit: hertz (Hz).
Relation:
ν = 1/T
8. What is angular frequency?
Answer:
Angular frequency is the rate of change of phase of an oscillating particle with time.
It is represented by ω.
ω = 2πν
or
ω = 2π/T
SI unit: radian per second (rad s⁻¹).
9. What is phase in simple harmonic motion?
Answer:
Phase describes the state of an oscillating particle at a particular instant of time.
For the displacement equation:
x = A cos(ωt + φ)
The quantity (ωt + φ) is called the phase.
It determines the displacement and direction of motion of the particle at a given time.
10. What is initial phase?
Answer:
Initial phase is the phase of an oscillating particle at time t = 0.
In the equation:
x = A cos(ωt + φ)
At t = 0, phase = φ.
Therefore, φ is called the initial phase or phase constant.
11. What is restoring force?
Answer:
Restoring force is the force that acts on an oscillating body towards its equilibrium or mean position.
It tries to bring the displaced body back to its original position.
For SHM:
F = −kx
The negative sign indicates that the force acts opposite to the displacement.
12. What is the relation between acceleration and displacement in SHM?
Answer:
In simple harmonic motion, acceleration is directly proportional to displacement and is always directed towards the mean position.
a ∝ −x
Therefore,
a = −ω²x
13. What is the relation between velocity and displacement in SHM?
Answer:
The velocity of a particle executing SHM is given by:
v = ±ω√(A² − x²)
where:
v = Velocity
ω = Angular frequency
A = Amplitude
x = Displacement
At the mean position, velocity is maximum.
At the extreme positions, velocity is zero.
14. What is the total mechanical energy of a particle executing SHM?
Answer:
The total mechanical energy of a particle executing ideal SHM is the sum of its kinetic energy and potential energy.
Total energy = Kinetic energy + Potential energy
E = (1/2)mv² + (1/2)mω²x²
For SHM:
E = (1/2)mω²A²
The total mechanical energy remains constant when there is no energy loss.
15. What is a simple pendulum?
Answer:
A simple pendulum consists of a small heavy bob suspended from a fixed support by a light, inextensible string.
When the bob is displaced slightly from its mean position and released, it oscillates to and fro.
For small angular oscillations, its time period is:
T = 2π√(l/g)
where:
l = Length of the pendulum
g = Acceleration due to gravity
EXERCISE QUESTIONS AND ANSWERS
Q13.1. Which of the following examples represent periodic motion?
(a) A swimmer completing one return trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its north-south direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Answer:
Periodic motion is motion that repeats itself after a fixed interval of time.
(a) A swimmer completing one return trip
A single return trip does not represent repeated motion at regular intervals. Therefore, it is not periodic motion.
(b) A freely suspended bar magnet displaced from its north-south direction and released
The magnet oscillates about its equilibrium direction. This motion repeats after equal intervals of time.
Therefore, it is periodic motion.
(c) A hydrogen molecule rotating about its centre of mass
The rotation of the molecule repeats after every complete revolution.
Therefore, it is periodic motion.
(d) An arrow released from a bow
The arrow moves forward and does not repeat its motion.
Therefore, it is not periodic motion.
Final Answer:
Examples (b) and (c) represent periodic motion.
Q13.2. Which of the following examples represent nearly simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) The rotation of the Earth about its axis.
(b) Motion of an oscillating mercury column in a U-tube.
(c) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowest point.
(d) General vibrations of a polyatomic molecule about its equilibrium position.
Answer:
(a) Rotation of the Earth about its axis
The Earth rotates periodically, but it does not move to and fro about a mean position.
Therefore, it is periodic but not SHM.
(b) Oscillation of a mercury column in a U-tube
When the mercury column is displaced, the restoring force is proportional to the displacement for small oscillations.
Therefore, it performs SHM.
(c) Ball bearing inside a smooth curved bowl
When the ball bearing is displaced slightly from the lowest point, a restoring force acts towards the lowest point.
For small displacements, the restoring force is approximately proportional to displacement.
Therefore, it performs nearly SHM.
(d) General vibrations of a polyatomic molecule
A polyatomic molecule can have several vibrational modes with different frequencies. Their combination does not generally represent one single SHM.
Final Answer:
Nearly SHM: (b) and (c).
Periodic but not SHM: (a) and (d).
Q13.3. Figure 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion in each case of periodic motion?
Answer:
A motion is periodic if its complete pattern repeats after a fixed interval of time.
(a) Plot showing displacement increasing steadily with time
The motion does not repeat itself.
Therefore, it is not periodic.
(b) Plot showing a repeating saw-tooth pattern
The complete pattern repeats after every 2 seconds.
Therefore, it is periodic.
Time period = 2 s.
(c) Plot showing irregular changes in displacement
The same displacement may occur at different times, but the complete motion does not repeat after a fixed interval.
Therefore, it is not periodic.
(d) Plot showing a repeating smooth wave
The complete pattern repeats after every 2 seconds.
Therefore, it is periodic.
Time period = 2 s.
Final Answer:
Plots (b) and (d) represent periodic motion.
Time period of each = 2 seconds.
Plots (a) and (c) are non-periodic.
Q13.4. Which of the following functions of time represent (a) simple harmonic motion, (b) periodic but not simple harmonic motion, and (c) non-periodic motion? Give the period for each case of periodic motion. Here ω is any positive constant.
(a) sin ωt − cos ωt
(b) sin³ ωt
(c) 3 cos (π/4 − 2ωt)
(d) cos ωt + cos 3ωt + cos 5ωt
(e) exp(−ω²t²)
(f) 1 + ωt + ω²t²
Answer:
(a) sin ωt − cos ωt
Using the identity:
sin A − cos A = √2 sin(A − π/4)
Therefore,
x = √2 sin(ωt − π/4)
This is the equation of SHM.
Time period:
T = 2π/ω
(b) sin³ ωt
Using the identity:
sin³ θ = (3 sin θ − sin 3θ)/4
Therefore,
sin³ ωt = (3 sin ωt − sin 3ωt)/4
It is a combination of two harmonic motions of different frequencies.
Therefore, it is periodic but not SHM.
Time period = 2π/ω.
(c) 3 cos (π/4 − 2ωt)
This can be written as:
x = 3 cos(2ωt − π/4)
This is the equation of SHM.
Angular frequency = 2ω.
Time period:
T = 2π/2ω
T = π/ω
(d) cos ωt + cos 3ωt + cos 5ωt
This is a combination of three harmonic functions having different frequencies.
It is periodic but not SHM.
The common period is:
T = 2π/ω
(e) exp(−ω²t²)
This function does not repeat itself after equal intervals of time.
Therefore, it is non-periodic.
(f) 1 + ωt + ω²t²
This function does not repeat itself after equal intervals of time.
Therefore, it is non-periodic.
Final Answer:
(a) SHM, T = 2π/ω.
(b) Periodic but not SHM, T = 2π/ω.
(c) SHM, T = π/ω.
(d) Periodic but not SHM, T = 2π/ω.
(e) Non-periodic.
(f) Non-periodic.
Q13.5. A particle is in linear simple harmonic motion between two points A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is:
(a) At the end A.
(b) At the end B.
(c) At the midpoint of AB going towards A.
(d) At 2 cm away from B going towards A.
(e) At 3 cm away from A going towards B.
(f) At 4 cm away from B going towards A.
Answer:
Distance between A and B = 10 cm.
The midpoint is the mean position O.
Taking O as origin:
A = −5 cm
B = +5 cm
The acceleration and restoring force are always directed towards the mean position.
(a) At the end A
At A, velocity is zero.
Acceleration and force act towards B, in the positive direction.
Therefore:
v = 0, a = +, F = +
(b) At the end B
At B, velocity is zero.
Acceleration and force act towards A, in the negative direction.
Therefore:
v = 0, a = −, F = −
(c) At the midpoint going towards A
At the midpoint, acceleration and force are zero.
The particle moves towards A, so velocity is negative.
Therefore:
v = −, a = 0, F = 0
(d) At 2 cm away from B going towards A
Position from mean point:
x = 5 − 2 = +3 cm
Velocity is negative because the particle moves towards A.
Acceleration and force act towards the mean position, also in the negative direction.
Therefore:
v = −, a = −, F = −
(e) At 3 cm away from A going towards B
Position from mean point:
x = −5 + 3 = −2 cm
Velocity is positive because the particle moves towards B.
Acceleration and force act towards the mean position, in the positive direction.
Therefore:
v = +, a = +, F = +
(f) At 4 cm away from B going towards A
Position from mean point:
x = 5 − 4 = +1 cm
Velocity is negative.
Acceleration and force act towards the mean position, also in the negative direction.
Therefore:
v = −, a = −, F = −
Final Answer:
| Position | Velocity | Acceleration | Force |
|---|---|---|---|
| (a) At A | 0 | + | + |
| (b) At B | 0 | − | − |
| (c) Midpoint towards A | − | 0 | 0 |
| (d) 2 cm from B towards A | − | − | − |
| (e) 3 cm from A towards B | + | + | + |
| (f) 4 cm from B towards A | − | − | − |
Q13.6. Which of the following relationships between acceleration a and displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = −200x²
(c) a = −10x
(d) a = 100x³
Answer:
The condition for simple harmonic motion is:
a = −ω²x
This means acceleration must be directly proportional to displacement and opposite in direction.
(a) a = 0.7x
Acceleration is in the same direction as displacement.
Therefore, it is not SHM.
(b) a = −200x²
Acceleration is proportional to the square of displacement, not displacement itself.
Therefore, it is not SHM.
(c) a = −10x
Acceleration is directly proportional to displacement and opposite in direction.
Comparing with:
a = −ω²x
ω² = 10
ω = √10 rad/s
Therefore, it represents SHM.
(d) a = 100x³
Acceleration is proportional to the cube of displacement.
Therefore, it is not SHM.
Final Answer:
Only option (c), a = −10x, represents simple harmonic motion.
Q13.7. The motion of a particle executing simple harmonic motion is described by the displacement function x(t) = A cos(ωt + φ). If the initial position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s⁻¹. If instead of the cosine function, we choose the sine function to describe the SHM, x = B sin(ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions?
Answer:
Given:
Initial displacement, x₀ = 1 cm
Initial velocity, v₀ = ω cm/s
Angular frequency, ω = π s⁻¹
Part 1: Using cosine function
x = A cos(ωt + φ)
At t = 0:
1 = A cos φ
Therefore,
A cos φ = 1 ...(1)
Velocity is:
v = −Aω sin(ωt + φ)
At t = 0:
ω = −Aω sin φ
Therefore,
A sin φ = −1 ...(2)
Squaring and adding equations (1) and (2):
A² cos²φ + A² sin²φ = 1 + 1
A² = 2
A = √2 cm
Now,
tan φ = −1
Since cos φ is positive and sin φ is negative:
φ = −π/4
Part 2: Using sine function
x = B sin(ωt + α)
At t = 0:
1 = B sin α ...(3)
Velocity:
v = Bω cos(ωt + α)
At t = 0:
ω = Bω cos α
B cos α = 1 ...(4)
Squaring and adding equations (3) and (4):
B² = 1 + 1
B = √2 cm
Also,
tan α = 1
Since both sine and cosine are positive:
α = π/4
Final Answer:
Using cosine function:
Amplitude = √2 cm
Initial phase = −π/4 rad
Using sine function:
Amplitude = √2 cm
Initial phase = π/4 rad
Q13.8. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Answer:
Given:
Maximum reading = 50 kg
Length of scale = 20 cm = 0.20 m
Time period, T = 0.6 s
g = 9.8 m/s²
Step 1: Calculate the spring constant.
At maximum reading:
Force = mg
F = 50 × 9.8
F = 490 N
Using Hooke's law:
F = kx
k = F/x
k = 490/0.20
k = 2450 N/m
Step 2: Calculate the mass of the body.
Time period of a spring-mass system:
T = 2π√(m/k)
Squaring both sides:
T² = 4π²m/k
Therefore,
m = kT²/4π²
Substituting:
m = (2450 × 0.6²)/(4 × 3.14²)
m ≈ 22.3 kg
Step 3: Calculate the weight.
Weight = mg
= 22.3 × 9.8
= 218.5 N
Final Answer:
Weight of the body = approximately 219 N.
Q13.9. A spring having a spring constant of 1200 N m⁻¹ is mounted on a horizontal table. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine:
(i) Frequency of oscillations.
(ii) Maximum acceleration of the mass.
(iii) Maximum speed of the mass.
Answer:
Given:
Spring constant, k = 1200 N/m
Mass, m = 3 kg
Amplitude, A = 2 cm = 0.02 m
(i) Frequency of oscillations
Angular frequency:
ω = √(k/m)
ω = √(1200/3)
ω = √400
ω = 20 rad/s
Frequency:
ν = ω/2π
ν = 20/(2 × 3.14)
ν ≈ 3.18 Hz
(ii) Maximum acceleration
Maximum acceleration:
amax = ω²A
amax = 20² × 0.02
amax = 400 × 0.02
amax = 8 m/s²
(iii) Maximum speed
Maximum speed:
vmax = ωA
vmax = 20 × 0.02
vmax = 0.4 m/s
Final Answer:
(i) Frequency = 3.18 Hz approximately.
(ii) Maximum acceleration = 8 m/s².
(iii) Maximum speed = 0.4 m/s.
Q13.10. In Exercise 13.9, let us take the position of the mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is:
(a) At the mean position.
(b) At the maximum stretched position.
(c) At the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, amplitude or initial phase?
Answer:
From Exercise 13.9:
Angular frequency, ω = 20 rad/s
Amplitude, A = 2 cm
The general displacement equation is:
x = A sin(ωt + φ)
or
x = A cos(ωt + φ)
(a) At the mean position
At t = 0, x = 0.
Assuming the mass initially moves towards the right:
x = A sin ωt
Substituting:
x = 2 sin 20t cm
(b) At the maximum stretched position
At t = 0, x = +2 cm.
Therefore, the displacement equation is:
x = A cos ωt
x = 2 cos 20t cm
(c) At the maximum compressed position
At t = 0, x = −2 cm.
Therefore,
x = −A cos ωt
x = −2 cos 20t cm
Final Answer:
(a) x = 2 sin 20t cm
(b) x = 2 cos 20t cm
(c) x = −2 cos 20t cm
All three equations have the same amplitude and frequency. They differ only in their initial phase.
Q13.11. Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated on each figure. Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P in each case:
(a) Radius = 3 cm, period = 2 s, P starts at the bottom of the circle on the negative y-axis and moves clockwise.
(b) Radius = 2 m, period = 4 s, P starts at the left end of the circle on the negative x-axis.
Answer:
The x-projection of uniform circular motion represents SHM.
The displacement equation is:
x = A cos(ωt + φ)
where:
ω = 2π/T
(a) First circular motion
Given:
Radius, A = 3 cm
Time period, T = 2 s
Angular frequency:
ω = 2π/2
ω = π rad/s
The particle starts at the bottom of the circle.
Therefore, initial x-displacement is zero.
As it moves clockwise, its x-projection initially moves towards the negative direction.
Hence,
x = −A sin ωt
Substituting:
x = −3 sin πt cm
(b) Second circular motion
Given:
Radius, A = 2 m
Time period, T = 4 s
Angular frequency:
ω = 2π/4
ω = π/2 rad/s
The particle starts at the left end of the circle.
Therefore, its initial x-displacement is −2 m.
Hence,
x = −A cos ωt
Substituting:
x = −2 cos(πt/2) m
Final Answer:
(a) x = −3 sin πt cm
(b) x = −2 cos(πt/2) m
Q13.12. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial position of the particle, the radius of the circle and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case. Here x is in cm and t is in seconds.
(a) x = −2 sin (3t + π/3)
(b) x = cos (π/6 − t)
(c) x = 3 sin (2πt + π/4)
(d) x = 2 cos πt
Answer:
For the reference circle:
Radius of circle = Amplitude of SHM.
Angular speed of circular motion = Angular frequency of SHM.
Initial position is determined by the initial phase angle.
We write each equation in the form:
x = A cos(ωt + φ)
(a) x = −2 sin(3t + π/3)
Using the identity:
−sin θ = cos(θ + π/2)
Therefore,
x = 2 cos(3t + π/3 + π/2)
x = 2 cos(3t + 5π/6)
Comparing with the standard equation:
Amplitude = 2 cm
Angular speed = 3 rad/s
Initial phase = 5π/6 rad
(b) x = cos(π/6 − t)
Using the property cos(−θ) = cos θ:
x = cos(t − π/6)
Amplitude = 1 cm
Angular speed = 1 rad/s
Initial phase = −π/6 rad
(c) x = 3 sin(2πt + π/4)
Using:
sin θ = cos(θ − π/2)
x = 3 cos(2πt + π/4 − π/2)
x = 3 cos(2πt − π/4)
Amplitude = 3 cm
Angular speed = 2π rad/s
Initial phase = −π/4 rad
(d) x = 2 cos πt
Comparing directly:
Amplitude = 2 cm
Angular speed = π rad/s
Initial phase = 0
Final Answer:
| Motion | Radius of circle | Angular speed | Initial phase |
|---|---|---|---|
| (a) | 2 cm | 3 rad/s | 5π/6 |
| (b) | 1 cm | 1 rad/s | −π/6 |
| (c) | 3 cm | 2π rad/s | −π/4 |
| (d) | 2 cm | π rad/s | 0 |
For each reference circle, draw the radius equal to the given amplitude and place the rotating particle at the angle specified by the initial phase.
Q13.13. Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 13.21(b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Figure 13.21(b) is stretched by the same force F.
(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in Figure (a) and the two masses in Figure (b) are released, what is the period of oscillation in each case?
Answer:
(a) Maximum extension of the spring
According to Hooke's law:
F = kx
Therefore,
x = F/k
In both cases, the tension in the spring is F.
Hence, maximum extension in both cases is:
x = F/k
(b) Time period of oscillation
Case (a): One end fixed and one mass attached
The time period of a spring-mass system is:
T = 2π√(m/k)
Case (b): Two equal masses attached to the two ends
The two masses move symmetrically in opposite directions about the centre of the spring.
The effective spring constant for the motion of either mass relative to the centre is 2k.
Therefore,
T = 2π√(m/2k)
Final Answer:
(a) Maximum extension in both cases = F/k
(b) Time period:
Case (a): T = 2π√(m/k)
Case (b): T = 2π√(m/2k)
Q13.14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
Answer:
Given:
Stroke = 1.0 m
Stroke = 2A
Angular frequency, ω = 200 rad/min
Step 1: Calculate amplitude.
2A = 1.0
A = 0.5 m
Step 2: Calculate maximum speed.
Maximum speed:
vmax = Aω
Substituting:
vmax = 0.5 × 200
vmax = 100 m/min
Converting into m/s:
vmax = 100/60
vmax = 1.67 m/s
Final Answer:
Maximum speed = 100 m/min or approximately 1.67 m/s.
Q13.15. The acceleration due to gravity on the surface of the Moon is 1.7 m s⁻². What is the time period of a simple pendulum on the surface of the Moon if its time period on the surface of the Earth is 3.5 s? Take the acceleration due to gravity on the surface of the Earth as 9.8 m s⁻².
Answer:
Given:
Time period on Earth, TE = 3.5 s
Gravity on Earth, gE = 9.8 m/s²
Gravity on Moon, gM = 1.7 m/s²
The time period of a simple pendulum is:
T = 2π√(l/g)
For the same pendulum length:
T ∝ 1/√g
Therefore,
TM/TE = √(gE/gM)
TM = TE × √(gE/gM)
Substituting:
TM = 3.5 × √(9.8/1.7)
TM = 3.5 × √5.765
TM = 3.5 × 2.401
TM ≈ 8.4 s
Final Answer:
Time period of the pendulum on the Moon = 8.4 seconds approximately.
Q13.16. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
Answer:
Given:
Length of pendulum = l
Mass of bob = M
Radius of circular track = R
Speed of car = v
Acceleration due to gravity = g
When the car moves along a circular track, the bob experiences an effective horizontal acceleration due to the circular motion.
Centrifugal acceleration:
ac = v²/R
The effective acceleration is the resultant of:
Vertical acceleration due to gravity, g.
Horizontal acceleration, v²/R.
Therefore,
geff = √[g² + (v²/R)²]
geff = √(g² + v⁴/R²)
The time period of a simple pendulum in an effective gravitational field is:
T = 2π√(l/geff)
Substituting:
T = 2π√[l/√(g² + v⁴/R²)]
Final Answer:
Time period:
T = 2π√[l/√(g² + v⁴/R²)]
The mass of the bob does not affect the time period.
Q13.17. A cylindrical piece of cork of density ρ, base area A and height h floats in a liquid of density ρ₁. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T = 2π√(hρ/ρ₁g). Ignore damping due to viscosity of the liquid.
Answer:
Given:
Density of cork = ρ
Base area = A
Height = h
Density of liquid = ρ₁
Acceleration due to gravity = g
Step 1: Mass of the cork
Volume of cork:
V = Ah
Mass:
m = Density × Volume
m = ρAh
Step 2: Restoring force
When the cork is pushed down by a small distance x, the additional volume immersed in the liquid is:
Additional volume = Ax
Additional buoyant force:
F = ρ₁Axg
This force acts upwards, opposite to the downward displacement.
Therefore, restoring force:
F = −ρ₁Agx
Comparing with Hooke's law:
F = −kx
We get:
k = ρ₁Ag
Step 3: Calculate the time period
For a spring-mass system:
T = 2π√(m/k)
Substituting:
T = 2π√(ρAh/ρ₁Ag)
Cancelling A:
T = 2π√(hρ/ρ₁g)
This is the required expression.
Conclusion:
The restoring force is directly proportional to displacement and acts in the opposite direction. Therefore, the cork executes simple harmonic motion.
Q13.18. One end of a U-tube containing mercury is connected to a suction pump and the other end to the atmosphere. A small pressure difference is maintained between the two columns. Show that when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Answer:
Consider a U-tube containing mercury.
When the suction pump is removed, both ends of the tube are exposed to atmospheric pressure.
Let:
A = Cross-sectional area of the tube
ρ = Density of mercury
L = Total length of the mercury column
y = Small displacement of the mercury column in one arm
Step 1: Difference in liquid levels
When the mercury column is displaced by y in one arm, it moves downward by y in the other arm.
Therefore, the difference in the levels of mercury is:
Difference = 2y
Step 2: Restoring force
The pressure difference between the two arms is:
ΔP = ρg(2y)
Force = Pressure × Area
F = ΔP × A
F = 2ρgAy
This force acts towards the equilibrium position.
Therefore,
F = −2ρgAy
Step 3: Calculate the mass of mercury
Volume of mercury:
V = AL
Mass:
m = ρAL
Step 4: Apply Newton's second law
F = ma
Therefore,
ma = −2ρgAy
Substituting m = ρAL:
ρALa = −2ρgAy
Cancelling ρA:
La = −2gy
Therefore,
a = −(2g/L)y
Comparing with the standard SHM equation:
a = −ω²y
We get:
ω² = 2g/L
ω = √(2g/L)
Step 5: Calculate the time period
T = 2π/ω
Therefore,
T = 2π√(L/2g)
Conclusion:
The acceleration is directly proportional to displacement and acts in the opposite direction.
Hence, the mercury column executes simple harmonic motion.
IMPORTANT QUESTIONS AND ANSWERS
1. Differentiate between simple harmonic motion and periodic motion.
Answer:
| Periodic Motion | Simple Harmonic Motion |
|---|---|
| Motion repeats after equal intervals of time. | A special type of oscillatory periodic motion. |
| Restoring force need not be proportional to displacement. | Restoring force is directly proportional to displacement and opposite in direction. |
| It may or may not be oscillatory. | It is oscillatory about a mean position. |
| Example: Uniform circular motion. | Example: Small oscillations of a spring-mass system. |
2. Derive the displacement equation of simple harmonic motion.
Answer:
In SHM, acceleration is:
a = −ω²x
We know:
a = d²x/dt²
Therefore,
d²x/dt² = −ω²x
Rearranging:
d²x/dt² + ω²x = 0
The solution of this differential equation is:
x = A cos(ωt + φ)
where:
A = Amplitude
ω = Angular frequency
t = Time
φ = Initial phase
This is the displacement equation of SHM.
3. Derive the expression for velocity in simple harmonic motion.
Answer:
The displacement equation is:
x = A cos(ωt + φ)
Differentiating with respect to time:
v = dx/dt
Therefore,
v = −Aω sin(ωt + φ)
We know:
x = A cos(ωt + φ)
Therefore,
cos(ωt + φ) = x/A
Using the identity:
sin²θ + cos²θ = 1
sin²(ωt + φ) = 1 − x²/A²
Therefore,
v² = A²ω²(1 − x²/A²)
v² = ω²(A² − x²)
Hence,
v = ±ω√(A² − x²)
This is the required expression.
4. Derive the expression for acceleration in simple harmonic motion.
Answer:
The displacement equation is:
x = A cos(ωt + φ)
Velocity:
v = −Aω sin(ωt + φ)
Differentiating again:
a = dv/dt
a = −Aω² cos(ωt + φ)
But:
x = A cos(ωt + φ)
Therefore,
a = −ω²x
This is the required expression for acceleration in SHM.
5. Derive the expression for time period of a spring-mass system.
Answer:
Consider a mass m attached to a spring of force constant k.
According to Hooke's law:
F = −kx
Using Newton's second law:
F = ma
Therefore,
ma = −kx
a = −(k/m)x
Comparing with the SHM equation:
a = −ω²x
We get:
ω² = k/m
ω = √(k/m)
We know:
T = 2π/ω
Therefore,
T = 2π√(m/k)
This is the required expression.
6. Derive the expression for the time period of a simple pendulum.
Answer:
Consider a simple pendulum of length l and bob mass m.
Let the bob be displaced through a small angle θ from its mean position.
The restoring force along the direction of motion is:
F = −mg sin θ
For small angles:
sin θ ≈ θ
Therefore,
F = −mgθ
For small angular displacement:
x = lθ
Thus,
θ = x/l
Substituting:
F = −mg(x/l)
F = −(mg/l)x
Using Newton's second law:
ma = −(mg/l)x
Cancelling m:
a = −(g/l)x
Comparing with:
a = −ω²x
We get:
ω² = g/l
ω = √(g/l)
Time period:
T = 2π/ω
Therefore,
T = 2π√(l/g)
This is the required expression for the time period of a simple pendulum.
7. Derive the expression for kinetic energy in SHM.
Answer:
Kinetic energy of a particle is:
K = (1/2)mv²
For SHM:
v² = ω²(A² − x²)
Substituting:
K = (1/2)mω²(A² − x²)
Therefore,
K = (1/2)mω²(A² − x²)
At the mean position, x = 0:
Kmax = (1/2)mω²A²
At the extreme positions, x = ±A:
K = 0
8. Derive the expression for potential energy in SHM.
Answer:
Restoring force:
F = −kx
Potential energy is equal to the work done against the restoring force.
Therefore,
U = ∫₀ˣ kx dx
U = kx²/2
For SHM:
k = mω²
Therefore,
U = (1/2)mω²x²
At the mean position, x = 0:
U = 0
At the extreme positions, x = ±A:
Umax = (1/2)mω²A²
9. Derive the expression for total energy in SHM.
Answer:
Total energy is the sum of kinetic energy and potential energy.
E = K + U
We know:
K = (1/2)mω²(A² − x²)
U = (1/2)mω²x²
Adding:
E = (1/2)mω²(A² − x²) + (1/2)mω²x²
E = (1/2)mω²A² − (1/2)mω²x² + (1/2)mω²x²
Therefore,
E = (1/2)mω²A²
The total energy remains constant throughout the motion in ideal SHM.
10. Explain the variation of kinetic energy and potential energy during SHM.
Answer:
During SHM, kinetic energy and potential energy continuously change into each other.
At the mean position:
Displacement = 0.
Velocity = Maximum.
Kinetic energy = Maximum.
Potential energy = Zero.
At the extreme positions:
Displacement = Maximum.
Velocity = Zero.
Kinetic energy = Zero.
Potential energy = Maximum.
At intermediate positions:
Both kinetic energy and potential energy are present.
Their sum remains constant.
Total energy:
E = (1/2)mω²A²
11. What is the relation between time period, frequency and angular frequency?
Answer:
Time period is the time taken to complete one oscillation.
Frequency is the number of oscillations completed in one second.
Therefore,
ν = 1/T
Angular frequency is:
ω = 2πν
Substituting:
ω = 2π/T
Hence,
T = 2π/ω
12. What are the characteristics of simple harmonic motion?
Answer:
The main characteristics of SHM are:
The motion is to and fro about a mean position.
The restoring force always acts towards the mean position.
Restoring force is directly proportional to displacement.
Acceleration is directly proportional to displacement.
Acceleration is opposite to displacement.
Velocity is maximum at the mean position.
Velocity is zero at the extreme positions.
Acceleration is zero at the mean position.
Acceleration is maximum at the extreme positions.
Total mechanical energy remains constant in ideal SHM.
IMPORTANT FORMULAE – CHAPTER 13
Frequency:
ν = 1/T
Time period:
T = 1/ν
Angular frequency:
ω = 2πν
Angular frequency in terms of time period:
ω = 2π/T
Displacement equation:
x = A cos(ωt + φ)
Alternative displacement equation:
x = A sin(ωt + φ)
Velocity:
v = −Aω sin(ωt + φ)
Velocity-displacement relation:
v = ±ω√(A² − x²)
Maximum velocity:
vmax = Aω
Acceleration:
a = −ω²x
Maximum acceleration:
amax = ω²A
Restoring force:
F = −mω²x
Force constant:
k = mω²
Time period of spring-mass system:
T = 2π√(m/k)
Angular frequency of spring-mass system:
ω = √(k/m)
Time period of simple pendulum:
T = 2π√(l/g)
Angular frequency of simple pendulum:
ω = √(g/l)
Kinetic energy:
K = (1/2)mω²(A² − x²)
Potential energy:
U = (1/2)mω²x²
Total energy:
E = (1/2)mω²A²
Time period of a floating cork:
T = 2π√(hρ/ρ₁g)
Time period of mercury in a U-tube:
T = 2π√(L/2g)
IMPORTANT SI UNITS
| Physical Quantity | SI Unit |
|---|---|
| Displacement | metre (m) |
| Amplitude | metre (m) |
| Time period | second (s) |
| Frequency | hertz (Hz) |
| Angular frequency | radian per second (rad/s) |
| Velocity | metre per second (m/s) |
| Acceleration | metre per second squared (m/s²) |
| Force | newton (N) |
| Spring constant | newton per metre (N/m) |
| Energy | joule (J) |
| Phase | radian (rad) |
IMPORTANT POINTS TO REMEMBER
Every simple harmonic motion is periodic, but every periodic motion is not SHM.
The mean position is the equilibrium position of an oscillator.
Amplitude is the maximum displacement from the mean position.
Velocity is maximum at the mean position.
Velocity is zero at the extreme positions.
Acceleration is zero at the mean position.
Acceleration is maximum at the extreme positions.
Restoring force always acts towards the mean position.
The time period of a spring-mass system depends on mass and spring constant.
The time period of a simple pendulum depends on its length and acceleration due to gravity.
For small oscillations, the time period of a simple pendulum is independent of the mass of its bob.
The total mechanical energy remains constant in ideal SHM.
Kinetic energy and potential energy continuously change into each other during SHM.
The x-projection of uniform circular motion represents simple harmonic motion.
The SI unit of frequency is hertz (Hz).
Comments
Post a Comment