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CodeVID-P11-13-SHM-01
SHM: Displacement, Velocity & Acceleration — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- 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. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The defining equation of SHM is:
- A.$a=\omega^2 x$
- B.$a=-\omega^2 x$
- C.$a=-\omega x$
- D.$a=\omega x^2$
2.
The SI unit of frequency is the:
- A.second
- B.hertz
- C.radian
- D.metre
3.
In SHM acceleration is maximum at the:
- A.mean position
- B.extreme position
- C.midpoint of mean and extreme
- D.all positions equally
4.
The phase difference between displacement and velocity in SHM is:
- A.$0$
- B.$\frac{\pi}{2}$
- C.$\pi$
- D.$2\pi$
5.
If the amplitude of an SHM is doubled (period unchanged), the maximum velocity becomes:
- A.halved
- B.unchanged
- C.doubled
- D.four times
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Distinguish between periodic and oscillatory motion with one example of each.
7.
A particle in SHM has $x=6\sin(2t)$ cm. Find its amplitude and time period.
8.
Write the expressions for velocity and acceleration of a particle in SHM in terms of displacement $x$.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A particle of amplitude 10 cm and period 2 s executes SHM. Find its velocity and acceleration when the displacement is 6 cm.
10.
Show that SHM can be regarded as the projection of uniform circular motion on a diameter.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Starting from $x=A\sin(\omega t+\phi)$, derive expressions for velocity and acceleration, and hence state where each is maximum and minimum.
Answer Key
Section A — Multiple Choice Questions
- (B) $a=-\omega^2 x$
- (B) hertz
- (B) extreme position
- (B) $\frac{\pi}{2}$
- (C) doubled
Section B — Short Answer (2 marks)
- Periodic motion repeats at fixed intervals (e.g. Earth around the Sun); oscillatory motion is to-and-fro about a mean position (e.g. a swinging pendulum). All oscillatory motion is periodic.
- $A=6\ \text{cm}$, $\omega=2\ \text{rad/s}$, $T=\frac{2\pi}{2}=\pi\approx3.14\ \text{s}$.
- $v=\omega\sqrt{A^2-x^2}$ and $a=-\omega^2 x$.
Section C — Short Answer (3 marks)
- $\omega=\pi\ \text{rad/s}$; $v=\pi\sqrt{10^2-6^2}=\pi\times8\approx25.1\ \text{cm/s}$; $a=-\omega^2 x=-\pi^2\times6\approx-59.2\ \text{cm/s}^2$.
- A reference particle moving on a circle of radius $A$ at angular speed $\omega$ has its foot of perpendicular on a diameter at $x=A\cos(\omega t)$ (or $A\sin$), whose acceleration is $-\omega^2 x$ — i.e. SHM.
Section D — Long Answer (5 marks)
- $v=\frac{dx}{dt}=A\omega\cos(\omega t+\phi)$ (max $A\omega$ at mean, zero at extremes); $a=\frac{dv}{dt}=-A\omega^2\sin(\omega t+\phi)=-\omega^2 x$ (zero at mean, max $\omega^2 A$ at extremes). Also $v=\omega\sqrt{A^2-x^2}$.
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