Vidaara.orgClass 11 · Physics
CodeVID-P11-13-SPR-01
Spring & Pendulum Systems — 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 period of a spring–mass system depends on:
- A.$g$
- B.mass and force constant
- C.amplitude only
- D.colour of the block
2.
For a simple pendulum, doubling the length changes the period by a factor of:
- A.$2$
- B.$\sqrt{2}$
- C.$\frac{1}{2}$
- D.$4$
3.
Two identical springs of constant $k$ in series give an effective constant of:
- A.$2k$
- B.$k$
- C.$\frac{k}{2}$
- D.$4k$
4.
In a damped oscillation the amplitude:
- A.stays constant
- B.increases with time
- C.decreases exponentially with time
- D.becomes infinite
5.
Resonance produces:
- A.minimum amplitude
- B.maximum amplitude
- C.zero frequency
- D.no oscillation
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
A 2 kg mass on a spring of constant $50\ \text{N/m}$ oscillates. Find its time period.
7.
Why is the time period of a simple pendulum independent of the mass of the bob?
8.
State the small-angle approximation used in deriving the pendulum period and when it is valid.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Derive the time period of a simple pendulum for small oscillations.
10.
Explain damped and forced oscillations, and define resonance.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Derive the expression for the time period of a spring–mass system and discuss the effect of connecting springs in series and in parallel.
Answer Key
Section A — Multiple Choice Questions
- (B) mass and force constant
- (B) $\sqrt{2}$
- (C) $\frac{k}{2}$
- (C) decreases exponentially with time
- (B) maximum amplitude
Section B — Short Answer (2 marks)
- $T=2\pi\sqrt{\frac{2}{50}}=2\pi\sqrt{0.04}=2\pi\times0.2\approx1.26\ \text{s}$.
- The restoring force $mg\sin\theta$ and inertia $m$ both scale with mass, so $m$ cancels; $T=2\pi\sqrt{L/g}$ has no mass term.
- $\sin\theta\approx\theta$ (with $\theta$ in radians), valid for small angles (roughly up to $10^\circ$).
Section C — Short Answer (3 marks)
- Restoring force $=-mg\sin\theta\approx -mg\theta=-\frac{mg}{L}x$, giving $\omega=\sqrt{g/L}$ and $T=2\pi\sqrt{L/g}$.
- Damped: amplitude decays due to a velocity-dependent force, $A=A_0 e^{-bt/2m}$. Forced: an external periodic force drives the system at its frequency. Resonance: amplitude is maximum when driving frequency equals the natural frequency.
Section D — Long Answer (5 marks)
- Hooke: $F=-kx=-m\omega^2 x\Rightarrow\omega=\sqrt{k/m}$, $T=2\pi\sqrt{m/k}$. Series: $\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}$ (softer, longer $T$). Parallel: $k_p=k_1+k_2$ (stiffer, shorter $T$).
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