Vidaara.orgClass 11 · Physics
CodeVID-P11-13-ENE-01
Energy in SHM — 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.
At the mean position of SHM, the energy is entirely:
- A.potential
- B.kinetic
- C.half kinetic, half potential
- D.zero
2.
The total energy of an SHM particle is proportional to:
- A.$A$
- B.$A^2$
- C.$\frac{1}{A}$
- D.$A^3$
3.
KE equals PE in SHM when the displacement is:
- A.$A$
- B.$\frac{A}{2}$
- C.$\frac{A}{\sqrt{2}}$
- D.$0$
4.
The potential energy of an SHM particle at the extreme position is:
- A.zero
- B.$\frac{1}{2}m\omega^2 A^2$
- C.$\frac{1}{4}m\omega^2 A^2$
- D.infinite
5.
The energy of an SHM particle varies with a frequency that is the frequency of the motion:
- A.equal to
- B.half
- C.twice
- D.four times
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Write the expressions for KE, PE and total energy of a particle in SHM at displacement $x$.
7.
A 1 kg body in SHM has $\omega=4\ \text{rad/s}$ and amplitude 0.1 m. Find its total energy.
8.
Why does the total energy of an SHM remain constant?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Show that the total energy of a particle in SHM is independent of its position.
10.
At displacement $x=\frac{A}{2}$, find the ratio of kinetic to potential energy of an SHM particle.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Derive expressions for the kinetic, potential and total energy of a particle in SHM and sketch how they vary with displacement.
Answer Key
Section A — Multiple Choice Questions
- (B) kinetic
- (B) $A^2$
- (C) $\frac{A}{\sqrt{2}}$
- (B) $\frac{1}{2}m\omega^2 A^2$
- (C) twice
Section B — Short Answer (2 marks)
- $KE=\frac{1}{2}m\omega^2(A^2-x^2)$, $PE=\frac{1}{2}m\omega^2 x^2$, $E=\frac{1}{2}m\omega^2 A^2$.
- $E=\frac{1}{2}\times1\times4^2\times0.1^2=\frac{1}{2}\times16\times0.01=0.08\ \text{J}$.
- The restoring force is conservative, so energy only transfers between KE and PE; their sum $\frac{1}{2}m\omega^2 A^2$ stays constant.
Section C — Short Answer (3 marks)
- $KE+PE=\frac{1}{2}m\omega^2(A^2-x^2)+\frac{1}{2}m\omega^2 x^2=\frac{1}{2}m\omega^2 A^2$; the $x$ terms cancel, so $E$ is constant.
- $\frac{KE}{PE}=\frac{A^2-x^2}{x^2}=\frac{A^2-A^2/4}{A^2/4}=\frac{3/4}{1/4}=3:1$.
Section D — Long Answer (5 marks)
- From $v=\omega\sqrt{A^2-x^2}$, $KE=\frac{1}{2}m\omega^2(A^2-x^2)$. Work against $F=-m\omega^2 x$ gives $PE=\frac{1}{2}m\omega^2 x^2$. Sum: $E=\frac{1}{2}m\omega^2 A^2$ (constant). KE is a downward parabola (max at $x=0$), PE an upward parabola (max at $x=\pm A$), and they add to the horizontal line $E$.
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