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CodeVID-P11-13-CH-01
Oscillations — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks. Show all working for Sections B, C and D.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
The defining relation of SHM is:
- A.$a=\omega^2 x$
- B.$a=-\omega^2 x$
- C.$a=-\omega x$
- D.$a=\omega x^2$
2.
In SHM the velocity is maximum at the:
- A.extreme position
- B.mean position
- C.point $x=A/2$
- D.point $x=A/\sqrt{2}$
3.
The total energy of an SHM particle is:
- A.maximum at the extreme
- B.maximum at the mean
- C.constant everywhere
- D.zero at the mean
4.
The period of a spring–mass system is independent of:
- A.mass
- B.force constant
- C.amplitude
- D.all of these
5.
The period of a simple pendulum depends on:
- A.mass of the bob
- B.amplitude
- C.length and $g$
- D.material of the bob
6.
Resonance corresponds to the amplitude being:
- A.minimum
- B.maximum
- C.zero
- D.constant
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
Define amplitude and time period of an oscillation.
8.
A particle in SHM has $x=4\sin(3t)$ cm. Find its maximum velocity.
9.
A 0.4 kg mass on a spring of constant $40\ \text{N/m}$ oscillates. Find the period.
10.
State the small-angle approximation used in deriving the simple pendulum formula.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
A particle of amplitude 5 cm and period 1 s executes SHM. Find its velocity and acceleration at displacement 3 cm.
12.
Show that the total energy of a particle in SHM is constant and equal to $\frac{1}{2}m\omega^2 A^2$.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Derive the time period of a simple pendulum for small oscillations and state two factors on which it does and does not depend.
14.
Discuss damped and forced oscillations and explain resonance with one real-life example.
Answer Key
Section A — Multiple Choice Questions
- (B) $a=-\omega^2 x$
- (B) mean position
- (C) constant everywhere
- (C) amplitude
- (C) length and $g$
- (B) maximum
Section B — Short Answer (2 marks)
- Amplitude is the maximum displacement from the mean position; time period is the time for one complete oscillation, $T=\frac{2\pi}{\omega}$.
- $A=4\ \text{cm}$, $\omega=3\ \text{rad/s}$, $v_{max}=A\omega=4\times3=12\ \text{cm/s}$.
- $T=2\pi\sqrt{\frac{0.4}{40}}=2\pi\sqrt{0.01}=2\pi\times0.1\approx0.63\ \text{s}$.
- $\sin\theta\approx\theta$ (in radians) for small angles, valid up to about $10^\circ$.
Section C — Short Answer (3 marks)
- $\omega=2\pi\ \text{rad/s}$; $v=2\pi\sqrt{5^2-3^2}=2\pi\times4\approx25.1\ \text{cm/s}$; $a=-\omega^2 x=-(2\pi)^2\times3\approx-118.4\ \text{cm/s}^2$.
- $KE+PE=\frac{1}{2}m\omega^2(A^2-x^2)+\frac{1}{2}m\omega^2 x^2=\frac{1}{2}m\omega^2 A^2$, independent of $x$.
Section D — Long Answer (5 marks)
- Restoring force $=-mg\sin\theta\approx -\frac{mg}{L}x$, so $\omega=\sqrt{g/L}$ and $T=2\pi\sqrt{L/g}$. It depends on $L$ and $g$; it does NOT depend on the mass of the bob or (for small angles) on the amplitude.
- Damped oscillations lose energy to friction so amplitude decays as $A_0 e^{-bt/2m}$. Forced oscillations are driven by an external periodic force at its own frequency. Resonance is the large-amplitude response when the driving frequency equals the natural frequency — e.g. pushing a swing in time with its natural rhythm builds a large amplitude.
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