Physics

Oscillations and Simple Harmonic Motion

138 Questions

Oscillations and simple harmonic motion focus on amplitude, damped vibrations, and force equations. These physics principles are essential for various engineering and civil services examinations. Review these problems to understand the core mechanics of oscillating bodies.

Damped harmonic oscillationSHM amplitudeForce equationsForced oscillationVibratory motion concepts

Oscillations and Simple Harmonic Motion Questions

Multiple choice
  1. a - 1, b - 3, c - 2

  2. a - 1, b - 2, c - 3

  3. a - 2, b - 3, c - 1

  4. a - 2, b - 1, c - 3

  5. a -3, b - 1, c - 2

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is the correct option. Amplitude of vibrations is the maximum displacement of a vibrating object from its central position. Time-period of vibrations is the time taken by a vibrating object to complete one vibration. Frequency of vibrations is the number of vibrations made in one second.

Multiple choice
  1. amplitude of the motion

  2. square of the amplitude of the motion

  3. cube of the amplititude of the motion

  4. square of the acceleration of the body

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In simple harmonic motion, total energy E = 1/2 k A^2 where k is force constant and A is amplitude. The energy is directly proportional to the SQUARE of amplitude, not amplitude itself or its cube. This is because both potential energy (at extremes) and kinetic energy (at mean position) depend on A^2.

Multiple choice
  1. a - 1, b - 3, c - 2

  2. a - 1, b - 2, c - 3

  3. a - 3, b - 1, c - 2

  4. a - 3, b - 2, c - 1

  5. a - 2, b - 3, c - 1

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is the correct option. Vibratory motion is a kind of oscillatory motion in which the moving object undergoes a change of shape or size. Non-periodic motion is a repetetitive motion which repeats itself, but not at fixed intervals of time. Non-uniform motion is the motion in which a body covers unequal distances in equal intervals of time along the same straight line.

Multiple choice
  1. $\frac{h}{2}\bigg(1-cos\frac{\pi \theta}{\beta}\bigg)$
  2. $\frac{\pi}{\beta}\frac{h}{2}sin\bigg(\frac{\pi \theta}{\beta}\bigg)$
  3. $\frac{\pi^2}{\beta^2}\frac{h}{2}cos\bigg(\frac{\pi \theta}{\beta}\bigg)$
  4. $-\frac{\pi^3}{\beta^3}\frac{h}{2}cos\bigg(\frac{\pi \theta}{\beta}\bigg)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

Any oscillation in which the amplitude of the oscillating quantity decreases with time is termed as

  1. Damped oscillation

  2. Free oscillation

  3. Depletion oscillation

  4. None of these

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Any oscillation in which the amplitude of the oscillating quantity decreases with time is termed as damped oscilaation

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

The amplitude of a damped oscillator becomes $\dfrac {1}{27}$ of initial value after $6\ minutes$. Its amplitude after $2\ minutes$ is:

  1. $\dfrac {A _{0}}{3}$
  2. $\dfrac {A _{0}}{9}$
  3. $\dfrac {A _{0}}{54}$
  4. $\dfrac {A _{0}}{81}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a damped oscillator, amplitude A(t) = A0 * e^(-bt). Given A(6) = A0/27, we have e^(-6b) = 1/27, so e^(-2b) = (1/27)^(1/3) = 1/3. Thus, A(2) = A0 * e^(-2b) = A0/3.

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

In damped vibrations, as time progresses, amplitude of oscillation

  1. decreases

  2. increases

  3. Remains same

  4. Data insufficient

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In damped oscillations, the relation of amplitude of oscillations with time is given by $y={ y } _{ o }{ e }^{ -bt }=\frac { { y } _{ o } }{ { e }^{ -bt } } $, where
${ y } _{ o }=$ initial amplitude of oscillation
$t=$time
$b=$damping constant
since $b>0\quad & \quad t>0;$
${ e }^{ bt }>1\ { \Rightarrow e }^{ -bt }<1\ { \Rightarrow { y } _{ o }e }^{ -bt }<{ y } _{ o }\ \Rightarrow y<{ y } _{ o }$
which means that as the time progress, its amplitude will decrease.

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

The amplitude of a damped oscilator becomes one-half after $t$ second. If the amplitude becomes $\dfrac {1}{n}$ after $3t$, second, then $n$ is equal to

  1. $\dfrac {1}{8}$
  2. $8$
  3. $\dfrac {1}{4}$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Amplitude A(t) = A0 * e^(-bt). Given A(t) = A0/2, e^(-bt) = 1/2. For 3t, A(3t) = A0 * (e^(-bt))^3 = A0 * (1/2)^3 = A0/8. Thus, n = 8.

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

Equation of motion for a particle performing damped harmonic oscillation is given as $x = e^{-1 t} cos (10 \pi t + \phi)$. The times when amplitude will half of the initial is :

  1. $27$
  2. $4$
  3. $1$
  4. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac{A _0}{2} = A _0 e^{-0.1t} \Rightarrow e^{-0.1t} = 2 \Rightarrow 0.1t = \ell n 2$
$t = \dfrac{\ell n 2}{0.1} = 10 \, \ell n2 \approx 6.93 \approx 7s$

Multiple choice physics free, damped and forced oscillations damped harmonic motion damped oscillation free, forced and damped oscillations

A particle is performing damped oscillation with frequency $5Hz$. After every $10$ oscillations its amplitude becomes half. find time from beginning after which the amplitude becomes $\dfrac{1}{1000}$ of its initial amplitude:

  1. $10 \,s$
  2. $20 \,s$
  3. $25 \,s$
  4. $50 \,s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$f = 5$
so $T = \dfrac{1}{5}$

$10T = \dfrac{10}{5} = 2$

$\dfrac{A _0}{1000} = A _0 \left(\dfrac{1}{2}\right)^{t/2}$

$(2)^{t/2} = 1000$

$\left(\dfrac{t}{2}\right) log 2 = 3$

$t = \dfrac{6}{log 2} \approx 20 s$