Tag: oscillatory motion

Quizzes (7) Questions (124)

Questions Related to oscillatory motion

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

The pendulum of a certain clock has time period $2.04 s$. How fast or slow does the clock run during $24$ hour?

  1. $28.8$ minutes slow
  2. $28.8$ minutes fast
  3. $14.4$ minutes fast
  4. $14.4$ minutes slow
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The clock loses time because its period is longer than the standard 2 seconds. The time lost in one day is (delta_T / T_actual) * 86400 seconds. With delta_T = 0.04s and T = 2s, the loss is (0.04/2.04) * 86400 approx 1694 seconds, which is about 28.2 minutes.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A bob is suspended from an ideal string of length $l$. Now it is pulled to a side through $60^{o}$ to vertical and rotates along a horizontal circle. Then its period of revolution is

  1. $ 2\pi \sqrt{ l/g }$
  2. $ \pi \sqrt{ l/2g }$
  3. $\pi\sqrt{ 2l/g }$
  4. $\pi\sqrt{ l/g }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a conical pendulum, the period is T = 2*pi*sqrt(h/g), where h = l*cos(theta). For theta = 60 degrees, cos(60) = 1/2, so h = l/2. Thus, T = 2*pi*sqrt(l/(2g)) = pi*sqrt(2l/g).

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Two simple pendulums have time period $4\ s$ and $5\ s$ respectively. If they started simultaneously from the mean positive in the same direction, then the phase difference between them by the time the larger one completes one osicillation is

  1. $\dfrac {\pi}{6}$
  2. $\dfrac {\pi}{3}$
  3. $\dfrac {\pi}{2}$
  4. $\dfrac {\pi}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The time periods are T1 = 4 s and T2 = 5 s. The larger one is T2 = 5 s, and in one complete oscillation of T2, time t = 5 s passes. The phase of the first pendulum in time t is omega_1 * t = (2*pi / T1) * t = (2*pi / 4) * 5 = 5*pi/2, which is 2*pi + pi/2, giving a phase of pi/2. The phase of the second pendulum is 2*pi, so the phase difference is pi/2 - 0 = pi/2.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Write the torque equation for the bob of a pendulum if it makes an angle of $\theta$ with the vertical and I is the moment of inertia of the bob w.r.t the point of suspension

  1. $I \dfrac{d^2 \theta}{dt^2}=mgL \cos \theta$
  2. $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$
  3. $I \dfrac{d^2 \theta}{dt^2}=mgL \tan \theta$
  4. $I \dfrac{d^2 \theta}{dt^2}=mg \sin \theta$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Taking the torque about the point of suspension, we can write $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$

The correct option is (b)

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

One end of spring of spring constant k is attached to the centre of a disc of mass m and radius R and the other end of the spring connected to a rigid wall. A string is wrapped on the disc and the end A of the string is pulled through a distance a and then released.
The disc is placed on a horizontal rough surface and there is no slipping at any contact point What is the amplitude of the oscillation of the centre of the disc?

  1. a

  2. 2a

  3. a/2

  4. none of these.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Displacement of the topmost point of the disc = a.
Disc undergoes rolling without slipping.
Hence the displacement of the centre of the disc = a/2
Thus the amplitude of the oscillation of the centre of the disc = a/2
Hence (C) is correct.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

The angular frequency of a torsional pendulum is $\omega$ rad/s. If the moment of inertia of the object is I, the torsional constant of the wire is related to the rotational kinetic energy of the disc, if the disc was rotating with an angular velocity $\omega$ is

  1. k= 2 KE

  2. k= KE

  3. k= 4 KE

  4. k= KE/2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

we know that $T=2 \pi \sqrt{I/k}$. Substituting the values given, we get, $k= I \omega^2 =2 \times $ kinetic energy

The correct option is (a)

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A small sphere is suspended by a string from the ceiling of a car. If the car begins to move with a constant acceleration $a$, the inclination of the string with the vertical is:-

  1. ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ in the direction of motion
  2. ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ opposite to the direction of motion
  3. ${\tan ^{ - 1}}\left( 2 \right)$ in the direction of motion
  4. ${\tan ^{ - 1}}\left( 2 \right)$ opposite to the direction of motion
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A simple pendulum with a metal bob has a time period $T$. Now the bob is immersed in a liquid which is non viscous. This time the time period is $4T$. The the ratio of densities of metal bpob and that of the liquid is

  1. $15:16$
  2. $16:15$
  3. $1:16$
  4. $16:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time period of a simple pendulum is given by T = 2*pi*sqrt(L/effective_g). When immersed in a non-viscous liquid, the effective acceleration due to gravity is g_eff = g * (1 - sigma/rho), where sigma is the density of the liquid and rho is the density of the metal bob. Since T' = 4T, squaring both sides gives 16 = rho / (rho - sigma), which simplifies to 16(rho - sigma) = rho, leading to 15*rho = 16*sigma, so rho/sigma = 16/15.