Chapter 14: Oscillations
Physics XI : Complete NCERT Exercise Solutions
Oscillations solved exercises. Learn SHM, simple pendulum, damping, resonance, energy in oscillations with step-by-step solutions.
Question 14.1
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 N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
(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 N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Answer & Explanation:
(a) Not periodic – The swimmer’s motion may not repeat in equal time intervals.
(b) Periodic – The magnet oscillates about its equilibrium position.
(c) Periodic – Rotation of the molecule is repetitive.
(d) Not periodic – The arrow’s motion is not repetitive; it stops after hitting the target.
Answer: (b) and (c) represent periodic motion.
(b) Periodic – The magnet oscillates about its equilibrium position.
(c) Periodic – Rotation of the molecule is repetitive.
(d) Not periodic – The arrow’s motion is not repetitive; it stops after hitting the target.
Answer: (b) and (c) represent periodic motion.
Question 14.2
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of 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 lower most point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
(a) the rotation of 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 lower most point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Answer & Explanation:
(a) Periodic but not SHM – Uniform circular motion, not to-and-fro about a mean position.
(b) Simple harmonic motion – Mercury column oscillates with restoring force proportional to displacement.
(c) Simple harmonic motion (for small displacements) – Motion under gravity in a bowl approximates SHM.
(d) Periodic but not SHM – Molecular vibrations are generally complex and not purely linear SHM.
Answer: (b) and (c) are SHM; (a) and (d) are periodic but not SHM.
(b) Simple harmonic motion – Mercury column oscillates with restoring force proportional to displacement.
(c) Simple harmonic motion (for small displacements) – Motion under gravity in a bowl approximates SHM.
(d) Periodic but not SHM – Molecular vibrations are generally complex and not purely linear SHM.
Answer: (b) and (c) are SHM; (a) and (d) are periodic but not SHM.
Question 14.3
Fig. 14.23 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 case of periodic motion)?
Answer & Explanation:
From the given plots (not shown here, but described in the textbook):
(a) Not periodic – Motion does not repeat.
(b) Periodic – Period = 2 s.
(c) Periodic – Period = 2 s.
(d) Not periodic – Motion does not repeat regularly.
Answer: Plots (b) and (c) are periodic with period 2 s each.
(a) Not periodic – Motion does not repeat.
(b) Periodic – Period = 2 s.
(c) Periodic – Period = 2 s.
(d) Not periodic – Motion does not repeat regularly.
Answer: Plots (b) and (c) are periodic with period 2 s each.
Question 14.4
Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (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
(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 & Explanation:
(a) SHM – Can be written as √2 sin(ωt – π/4). Period = 2π/ω.
(b) Periodic but not SHM – sin3 ωt is periodic but not sinusoidal. Period = π/ω.
(c) SHM – 3 cos(π/4 – 2ωt) = 3 cos(2ωt – π/4). Period = π/ω.
(d) Periodic but not SHM – Sum of cosines with different frequencies. Period = 2π/ω (LCM of periods).
(e) Non-periodic – Exponential decay, never repeats.
(f) Non-periodic – Polynomial function, not repetitive.
Summary:
SHM: (a), (c)
Periodic but not SHM: (b), (d)
Non-periodic: (e), (f)
(b) Periodic but not SHM – sin3 ωt is periodic but not sinusoidal. Period = π/ω.
(c) SHM – 3 cos(π/4 – 2ωt) = 3 cos(2ωt – π/4). Period = π/ω.
(d) Periodic but not SHM – Sum of cosines with different frequencies. Period = 2π/ω (LCM of periods).
(e) Non-periodic – Exponential decay, never repeats.
(f) Non-periodic – Polynomial function, not repetitive.
Summary:
SHM: (a), (c)
Periodic but not SHM: (b), (d)
Non-periodic: (e), (f)
Question 14.5
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 mid-point 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, and
(f) at 4 cm away from B going towards A.
(a) at the end A,
(b) at the end B,
(c) at the mid-point 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, and
(f) at 4 cm away from B going towards A.
Answer & Explanation:
Let A be at x = –5 cm, B at x = +5 cm, mean at x = 0.
(a) At A (x = –5 cm): v = 0, a = + (towards mean), F = +.
(b) At B (x = +5 cm): v = 0, a = –, F = –.
(c) At midpoint (x = 0) moving towards A: v = –, a = 0, F = 0.
(d) 2 cm from B (x = +3 cm) moving towards A: v = –, a = –, F = –.
(e) 3 cm from A (x = –2 cm) moving towards B: v = +, a = +, F = +.
(f) 4 cm from B (x = +1 cm) moving towards A: v = –, a = –, F = –.
Note: Acceleration and force are always directed towards mean position.
(a) At A (x = –5 cm): v = 0, a = + (towards mean), F = +.
(b) At B (x = +5 cm): v = 0, a = –, F = –.
(c) At midpoint (x = 0) moving towards A: v = –, a = 0, F = 0.
(d) 2 cm from B (x = +3 cm) moving towards A: v = –, a = –, F = –.
(e) 3 cm from A (x = –2 cm) moving towards B: v = +, a = +, F = +.
(f) 4 cm from B (x = +1 cm) moving towards A: v = –, a = –, F = –.
Note: Acceleration and force are always directed towards mean position.
Question 14.6
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = –200x2
(c) a = –10x
(d) a = 100x3
(a) a = 0.7x
(b) a = –200x2
(c) a = –10x
(d) a = 100x3
Answer & Explanation:
For SHM: a ∝ –x (acceleration proportional to displacement and directed opposite).
(a) a = 0.7x → Not SHM (same direction, not opposite).
(b) a = –200x2 → Not SHM (depends on x2, not linear).
(c) a = –10x → SHM (linear, opposite direction).
(d) a = 100x3 → Not SHM (non-linear).
Answer: Only (c) represents SHM.
(a) a = 0.7x → Not SHM (same direction, not opposite).
(b) a = –200x2 → Not SHM (depends on x2, not linear).
(c) a = –10x → SHM (linear, opposite direction).
(d) a = 100x3 → Not SHM (non-linear).
Answer: Only (c) represents SHM.
Question 14.7
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos(ωt + φ). If the initial (t = 0) 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 & Explanation:
Given: ω = π s–1, x(0) = 1 cm, v(0) = ω = π cm/s.
For x = A cos(ωt + φ):
At t = 0: x = A cos φ = 1 …(1)
v = –Aω sin φ = π …(2)
From (1) & (2): A2 = 12 + (π/π)2 = 2 ⇒ A = √2 cm.
tan φ = – (v/ωx) = – (π/(π×1)) = –1 ⇒ φ = –π/4 or 7π/4.
For x = B sin(ωt + α):
At t = 0: x = B sin α = 1 …(3)
v = Bω cos α = π …(4)
B2 = 12 + (π/π)2 = 2 ⇒ B = √2 cm.
tan α = xω/v = (1×π)/π = 1 ⇒ α = π/4 or 5π/4.
Answers:
Cosine form: A = √2 cm, φ = –π/4
Sine form: B = √2 cm, α = π/4
For x = A cos(ωt + φ):
At t = 0: x = A cos φ = 1 …(1)
v = –Aω sin φ = π …(2)
From (1) & (2): A2 = 12 + (π/π)2 = 2 ⇒ A = √2 cm.
tan φ = – (v/ωx) = – (π/(π×1)) = –1 ⇒ φ = –π/4 or 7π/4.
For x = B sin(ωt + α):
At t = 0: x = B sin α = 1 …(3)
v = Bω cos α = π …(4)
B2 = 12 + (π/π)2 = 2 ⇒ B = √2 cm.
tan α = xω/v = (1×π)/π = 1 ⇒ α = π/4 or 5π/4.
Answers:
Cosine form: A = √2 cm, φ = –π/4
Sine form: B = √2 cm, α = π/4
Question 14.8
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 & Explanation:
Maximum force = 50 kg × g = 50g N.
Spring extension for this force = 20 cm = 0.2 m.
Spring constant: k = F/x = 50g/0.2 = 250g N/m.
Period T = 2π√(m/k) = 0.6 s.
⇒ m/k = (T/2π)2 = (0.6/2π)2.
m = k × (0.6/2π)2 = 250g × (0.09/π2).
Weight = mg = 250g2 × (0.09/π2).
Taking g = 9.8 m/s2: Weight ≈ 22.3 N ≈ 2.27 kg.
Answer: About 22.3 N (or mass ≈ 2.27 kg).
Spring extension for this force = 20 cm = 0.2 m.
Spring constant: k = F/x = 50g/0.2 = 250g N/m.
Period T = 2π√(m/k) = 0.6 s.
⇒ m/k = (T/2π)2 = (0.6/2π)2.
m = k × (0.6/2π)2 = 250g × (0.09/π2).
Weight = mg = 250g2 × (0.09/π2).
Taking g = 9.8 m/s2: Weight ≈ 22.3 N ≈ 2.27 kg.
Answer: About 22.3 N (or mass ≈ 2.27 kg).
Question 14.9
A spring having a spring constant 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) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.
Answer & Explanation:
Given: k = 1200 N/m, m = 3 kg, A = 0.02 m.
(i) Angular frequency: ω = √(k/m) = √(1200/3) = 20 rad/s.
Frequency: ν = ω/2π = 20/(2π) ≈ 3.18 Hz.
(ii) Maximum acceleration: amax = ω2A = (20)2 × 0.02 = 8 m/s2.
(iii) Maximum speed: vmax = ωA = 20 × 0.02 = 0.4 m/s.
Answer: (i) ≈ 3.18 Hz, (ii) 8 m/s2, (iii) 0.4 m/s.
(i) Angular frequency: ω = √(k/m) = √(1200/3) = 20 rad/s.
Frequency: ν = ω/2π = 20/(2π) ≈ 3.18 Hz.
(ii) Maximum acceleration: amax = ω2A = (20)2 × 0.02 = 8 m/s2.
(iii) Maximum speed: vmax = ωA = 20 × 0.02 = 0.4 m/s.
Answer: (i) ≈ 3.18 Hz, (ii) 8 m/s2, (iii) 0.4 m/s.
Question 14.10
In Exercise 14.9, let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of 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, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer & Explanation:
ω = 20 rad/s, A = 0.02 m.
(a) At mean position (x=0) moving right: x = A sin ωt = 0.02 sin 20t.
(b) At maximum stretched (x=+A): x = A cos ωt = 0.02 cos 20t.
(c) At maximum compressed (x=–A): x = –A cos ωt = –0.02 cos 20t.
Differences: All have same frequency and amplitude. They differ only in initial phase.
(a) Phase = 0 for sine form.
(b) Phase = 0 for cosine form.
(c) Phase = π for cosine form.
(a) At mean position (x=0) moving right: x = A sin ωt = 0.02 sin 20t.
(b) At maximum stretched (x=+A): x = A cos ωt = 0.02 cos 20t.
(c) At maximum compressed (x=–A): x = –A cos ωt = –0.02 cos 20t.
Differences: All have same frequency and amplitude. They differ only in initial phase.
(a) Phase = 0 for sine form.
(b) Phase = 0 for cosine form.
(c) Phase = π for cosine form.
Question 14.11
Figures 14.25 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.
Answer & Explanation:
Case (a): Radius = 3 cm, T = 2 s, starts from (0,–3) moving clockwise.
ω = 2π/T = π rad/s.
Initial phase: At t=0, particle at negative y-axis, projection on x-axis = 0 moving positive.
x(t) = 3 sin(πt) cm.
Case (b): Radius = 2 m, T = 4 s, starts from (1,0) moving anticlockwise.
ω = π/2 rad/s.
Initial angle: cos–1(1/2) = π/3.
x(t) = 2 cos(πt/2 + π/3) m.
Answer:
(a) x = 3 sin(πt) cm
(b) x = 2 cos(πt/2 + π/3) m
ω = 2π/T = π rad/s.
Initial phase: At t=0, particle at negative y-axis, projection on x-axis = 0 moving positive.
x(t) = 3 sin(πt) cm.
Case (b): Radius = 2 m, T = 4 s, starts from (1,0) moving anticlockwise.
ω = π/2 rad/s.
Initial angle: cos–1(1/2) = π/3.
x(t) = 2 cos(πt/2 + π/3) m.
Answer:
(a) x = 3 sin(πt) cm
(b) x = 2 cos(πt/2 + π/3) m
Question 14.12
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) 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: (x is in cm and t is in s).
(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
(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 & Explanation:
Rewrite each in standard form x = A cos(ωt + φ) or A sin(ωt + φ):
(a) x = –2 sin(3t + π/3) = 2 cos(3t + 5π/6).
A = 2 cm, ω = 3 rad/s, φ = 5π/6.
(b) x = cos(π/6 – t) = cos(t – π/6).
A = 1 cm, ω = 1 rad/s, φ = –π/6.
(c) x = 3 sin(2πt + π/4) = 3 cos(2πt – π/4).
A = 3 cm, ω = 2π rad/s, φ = –π/4.
(d) x = 2 cos πt.
A = 2 cm, ω = π rad/s, φ = 0.
For each: Radius = A, angular speed = ω, initial phase = φ.
(a) x = –2 sin(3t + π/3) = 2 cos(3t + 5π/6).
A = 2 cm, ω = 3 rad/s, φ = 5π/6.
(b) x = cos(π/6 – t) = cos(t – π/6).
A = 1 cm, ω = 1 rad/s, φ = –π/6.
(c) x = 3 sin(2πt + π/4) = 3 cos(2πt – π/4).
A = 3 cm, ω = 2π rad/s, φ = –π/4.
(d) x = 2 cos πt.
A = 2 cm, ω = π rad/s, φ = 0.
For each: Radius = A, angular speed = ω, initial phase = φ.
Question 14.13
Figure 14.26 (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 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.26(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 Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
Answer & Explanation:
(a) Maximum extension:
Case (a): Extension = F/k.
Case (b): For spring with both ends pulled by F, total extension = F/k (same as case a).
(b) Period:
Case (a): T = 2π√(m/k).
Case (b): Each mass oscillates as if attached to half-spring of stiffness 2k. T = 2π√(m/(2k)).
Answers:
(a) Max extension = F/k for both.
(b) Ta = 2π√(m/k), Tb = 2π√(m/(2k)).
Case (a): Extension = F/k.
Case (b): For spring with both ends pulled by F, total extension = F/k (same as case a).
(b) Period:
Case (a): T = 2π√(m/k).
Case (b): Each mass oscillates as if attached to half-spring of stiffness 2k. T = 2π√(m/(2k)).
Answers:
(a) Max extension = F/k for both.
(b) Ta = 2π√(m/k), Tb = 2π√(m/(2k)).
Question 14.14
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 & Explanation:
Stroke = 2A = 1.0 m ⇒ A = 0.5 m.
ω = 200 rad/min = 200/60 rad/s = 10/3 rad/s.
Maximum speed vmax = ωA = (10/3) × 0.5 = 5/3 ≈ 1.667 m/s.
Answer: ≈ 1.67 m/s.
ω = 200 rad/min = 200/60 rad/s = 10/3 rad/s.
Maximum speed vmax = ωA = (10/3) × 0.5 = 5/3 ≈ 1.667 m/s.
Answer: ≈ 1.67 m/s.
Question 14.15
The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s–2)
Answer & Explanation:
T ∝ 1/√g.
Tmoon = Tearth × √(gearth/gmoon)
= 3.5 × √(9.8/1.7)
= 3.5 × √(5.7647)
≈ 3.5 × 2.4 = 8.4 s.
Answer: ≈ 8.4 s.
Tmoon = Tearth × √(gearth/gmoon)
= 3.5 × √(9.8/1.7)
= 3.5 × √(5.7647)
≈ 3.5 × 2.4 = 8.4 s.
Answer: ≈ 8.4 s.
Question 14.16
Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle: T = 2π√(m/k). A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π√(l/g). Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle: T = 2π√(m/k). A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π√(l/g). Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Answer & Explanation:
(a) For pendulum, restoring force = mg sinθ ≈ mgθ, effective k = mg/l. So T = 2π√(m/(mg/l)) = 2π√(l/g), independent of m.
(b) For larger θ, sinθ < θ, so restoring torque is less, motion slower, period increases.
(c) Yes, wristwatch works on internal mechanism, independent of free fall.
(d) In free fall, effective g = 0, so frequency = 0 (pendulum does not oscillate).
Answers:
(a) Because m cancels out in k = mg/l.
(b) Larger θ reduces effective restoring force, increasing T.
(c) Yes.
(d) Zero.
(b) For larger θ, sinθ < θ, so restoring torque is less, motion slower, period increases.
(c) Yes, wristwatch works on internal mechanism, independent of free fall.
(d) In free fall, effective g = 0, so frequency = 0 (pendulum does not oscillate).
Answers:
(a) Because m cancels out in k = mg/l.
(b) Larger θ reduces effective restoring force, increasing T.
(c) Yes.
(d) Zero.
Question 14.17
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 & Explanation:
In the car’s frame, bob experiences centrifugal force Mv2/R outward.
Effective acceleration g' = √(g2 + (v2/R)2).
For small oscillations in the radial direction (horizontal), restoring force comes from centrifugal variation.
Time period T = 2π√(l/(v2/R)) = 2π√(lR/v2).
Answer: T = 2π√(lR/v2).
Effective acceleration g' = √(g2 + (v2/R)2).
For small oscillations in the radial direction (horizontal), restoring force comes from centrifugal variation.
Time period T = 2π√(l/(v2/R)) = 2π√(lR/v2).
Answer: T = 2π√(lR/v2).
Question 14.18
A cylindrical piece of cork of density ρ of base area A and height h floats in a liquid of density ρl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T = 2π√(hρ/(ρl g)) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer & Explanation:
Let cork be depressed by x.
Extra buoyant force = – (ρl g A x).
Mass of cork = ρ A h.
Equation: ρ A h d2x/dt2 = – ρl g A x.
⇒ d2x/dt2 = – (ρl g/(ρ h)) x.
This is SHM with ω2 = ρl g/(ρ h).
Period T = 2π/ω = 2π√(ρ h/(ρl g)).
Proved.
Extra buoyant force = – (ρl g A x).
Mass of cork = ρ A h.
Equation: ρ A h d2x/dt2 = – ρl g A x.
⇒ d2x/dt2 = – (ρl g/(ρ h)) x.
This is SHM with ω2 = ρl g/(ρ h).
Period T = 2π/ω = 2π√(ρ h/(ρl g)).
Proved.
Question 14.19
One end of a U-tube containing mercury is connected to a suction pump and the other end to 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 & Explanation:
Let liquid be depressed by x in one limb, rises x in other.
Difference in height = 2x.
Restoring force = – (ρ g × 2x) × A, where A = cross-section area.
Mass of liquid = ρ × (2L) × A, where L = length of liquid column? Actually, mass oscillating = ρ × A × total length? For U-tube, total mass = ρ A Ltotal.
Equation: (ρ A L) d2x/dt2 = – 2ρ g A x.
⇒ d2x/dt2 = – (2g/L) x.
This is SHM with ω2 = 2g/L, where L = length of mercury column.
Proved.
Difference in height = 2x.
Restoring force = – (ρ g × 2x) × A, where A = cross-section area.
Mass of liquid = ρ × (2L) × A, where L = length of liquid column? Actually, mass oscillating = ρ × A × total length? For U-tube, total mass = ρ A Ltotal.
Equation: (ρ A L) d2x/dt2 = – 2ρ g A x.
⇒ d2x/dt2 = – (2g/L) x.
This is SHM with ω2 = 2g/L, where L = length of mercury column.
Proved.
Question 14.20
An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.
Answer & Explanation:
Let ball be pressed down by x.
Volume decreases by ΔV = a x.
For isothermal process: PV = constant.
Excess pressure ΔP = –P (ΔV/V) = –P (a x/V).
Restoring force = ΔP × a = – (P a2/V) x.
Equation: m d2x/dt2 = – (P a2/V) x.
⇒ SHM with ω2 = P a2/(m V).
Period T = 2π √(m V/(P a2)).
Proved.
Volume decreases by ΔV = a x.
For isothermal process: PV = constant.
Excess pressure ΔP = –P (ΔV/V) = –P (a x/V).
Restoring force = ΔP × a = – (P a2/V) x.
Equation: m d2x/dt2 = – (P a2/V) x.
⇒ SHM with ω2 = P a2/(m V).
Period T = 2π √(m V/(P a2)).
Proved.
Question 14.21
This exercise set covers fundamental concepts of oscillatory motion including simple harmonic motion, energy in SHM, simple pendulum, damped and forced oscillations, and resonance. Students will learn to solve problems involving displacement, velocity, acceleration, time period, and energy in oscillating systems. These questions strengthen conceptual understanding and problem-solving skills for board exams and competitive tests like JEE and NEET.
Answer & Explanation:
(a) For one wheel: mass m = 750 kg.
Static compression x0 = 0.15 m.
k = mg/x0 = (750×9.8)/0.15 ≈ 49000 N/m.
(b) Amplitude decreases by 50% in one period T.
A(t) = A0 e–bt/2m.
After t = T, A/A0 = 0.5 = e–bT/2m.
⇒ bT/2m = ln 2 ≈ 0.693.
T = 2π√(m/k) = 2π√(750/49000) ≈ 0.78 s.
b = (0.693 × 2m)/T ≈ (0.693×2×750)/0.78 ≈ 1330 kg/s.
Answers: (a) k ≈ 4.9×104 N/m, (b) b ≈ 1330 kg/s.
Static compression x0 = 0.15 m.
k = mg/x0 = (750×9.8)/0.15 ≈ 49000 N/m.
(b) Amplitude decreases by 50% in one period T.
A(t) = A0 e–bt/2m.
After t = T, A/A0 = 0.5 = e–bT/2m.
⇒ bT/2m = ln 2 ≈ 0.693.
T = 2π√(m/k) = 2π√(750/49000) ≈ 0.78 s.
b = (0.693 × 2m)/T ≈ (0.693×2×750)/0.78 ≈ 1330 kg/s.
Answers: (a) k ≈ 4.9×104 N/m, (b) b ≈ 1330 kg/s.
Question 14.22
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Answer & Explanation:
For SHM: x = A cos(ωt+φ), v = –Aω sin(ωt+φ).
K = ½ m v2 = ½ m A2 ω2 sin2(ωt+φ).
U = ½ k x2 = ½ m ω2 A2 cos2(ωt+φ).
Average over one period T:
<sin2> = <cos2> = ½.
So <K> = ½ m ω2 A2 × ½ = ¼ m ω2 A2.
<U> = ½ m ω2 A2 × ½ = ¼ m ω2 A2.
Hence <K> = <U>.
Proved.
K = ½ m v2 = ½ m A2 ω2 sin2(ωt+φ).
U = ½ k x2 = ½ m ω2 A2 cos2(ωt+φ).
Average over one period T:
<sin2> = <cos2> = ½.
So <K> = ½ m ω2 A2 × ½ = ¼ m ω2 A2.
<U> = ½ m ω2 A2 × ½ = ¼ m ω2 A2.
Hence <K> = <U>.
Proved.
Question 14.23
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist).
Answer & Explanation:
Moment of inertia of disc: I = ½ M R2 = ½ × 10 × (0.15)2 = 0.1125 kg m2.
For torsional oscillations: T = 2π√(I/α).
⇒ α = 4π2 I / T2.
= 4π2 × 0.1125 / (1.5)2
≈ (4×9.87×0.1125) / 2.25
≈ 4.44 / 2.25 ≈ 1.97 N m/rad.
Answer: α ≈ 1.97 N m/rad.
For torsional oscillations: T = 2π√(I/α).
⇒ α = 4π2 I / T2.
= 4π2 × 0.1125 / (1.5)2
≈ (4×9.87×0.1125) / 2.25
≈ 4.44 / 2.25 ≈ 1.97 N m/rad.
Answer: α ≈ 1.97 N m/rad.
Question 14.24
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.
Answer & Explanation:
A = 0.05 m, T = 0.2 s, ω = 2π/T = 10π rad/s.
Velocity: v = ω√(A2 – x2)
Acceleration: a = –ω2 x.
(a) x = 0.05 m:
v = 0, a = –(10π)2×0.05 = –100π2×0.05 ≈ –49.3 m/s2.
(b) x = 0.03 m:
v = 10π √(0.052 – 0.032) = 10π √(0.0016) = 10π × 0.04 = 0.4π ≈ 1.26 m/s.
a = –(10π)2×0.03 = –100π2×0.03 ≈ –29.6 m/s2.
(c) x = 0:
v = 10π × 0.05 = 0.5π ≈ 1.57 m/s.
a = 0.
Answers:
(a) v=0, a≈ –49.3 m/s2
(b) v≈1.26 m/s, a≈ –29.6 m/s2
(c) v≈1.57 m/s, a=0
Velocity: v = ω√(A2 – x2)
Acceleration: a = –ω2 x.
(a) x = 0.05 m:
v = 0, a = –(10π)2×0.05 = –100π2×0.05 ≈ –49.3 m/s2.
(b) x = 0.03 m:
v = 10π √(0.052 – 0.032) = 10π √(0.0016) = 10π × 0.04 = 0.4π ≈ 1.26 m/s.
a = –(10π)2×0.03 = –100π2×0.03 ≈ –29.6 m/s2.
(c) x = 0:
v = 10π × 0.05 = 0.5π ≈ 1.57 m/s.
a = 0.
Answers:
(a) v=0, a≈ –49.3 m/s2
(b) v≈1.26 m/s, a≈ –29.6 m/s2
(c) v≈1.57 m/s, a=0
Question 14.25
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0.
Answer & Explanation:
Let x = A cos(ωt + φ).
At t=0: x0 = A cos φ …(1)
v0 = –Aω sin φ …(2) (since pushed towards centre, velocity negative if x0 positive).
Squaring and adding: A2 = x02 + (v0/ω)2.
So amplitude A = √[x02 + (v0/ω)2].
Answer: A = √(x02 + (v0/ω)2).
At t=0: x0 = A cos φ …(1)
v0 = –Aω sin φ …(2) (since pushed towards centre, velocity negative if x0 positive).
Squaring and adding: A2 = x02 + (v0/ω)2.
So amplitude A = √[x02 + (v0/ω)2].
Answer: A = √(x02 + (v0/ω)2).
📘 Exam Preparation Tip:
This exercise set covers fundamental concepts of oscillatory motion including simple harmonic motion, energy in SHM, simple pendulum, damped and forced oscillations, and resonance. Students will learn to solve problems involving displacement, velocity, acceleration, time period, and energy in oscillating systems. These questions strengthen conceptual understanding and problem-solving skills for board exams and competitive tests like JEE and NEET.