Tag: angular simple harmonic motion

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Questions Related to angular simple harmonic motion

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.

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

A pendulum clock keeping correct time is taken to high altitudes,

  1. it will keep correct time

  2. its length should be increased to keep correct time

  3. its length should be decreased to keep correct time

  4. it cannot keep correct time even if the length is
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At high altitudes, g decreases. Since T = 2*pi*sqrt(l/g), T increases, meaning the clock runs slow. To keep correct time, T must be decreased, which requires decreasing the length l.

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

Restoring force on the bob of a simple pendulum of mass $100\ gm$ when its amplitude is ${ 1 }^{ 0 } $ is 

  1. $0.017\ N$
  2. $1.7\ N$
  3. $0.17\ N$
  4. $0.034\ N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The restoring force for a simple pendulum is F = mg*sin(theta). For theta = 1 degree, F = 0.1 * 9.8 * sin(1 degree) approx 0.1 * 9.8 * 0.01745 = 0.0171 N.

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

Simple pendulum of large length is made equal to the radius of earth. Its period of oscillation will be then?

  1. 83.5 minutes

  2. 59.8 minutes

  3. 42.3 minutes
    1. 15 minutes
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The time period of simple pendulum is given by:

$T=2\pi \sqrt{\dfrac{L}{g}}$

When length of pendulum is equal to the radius of earth. $R=L=6371\,\,Km=6371\times {{10}^{3}}\,Km$

So, time is

$ T=2\pi \sqrt{\dfrac{6371\times {{10}^{3}}}{10}} $

$ T=2\pi \times 798.18 $

$ T=5012.604\,seconds $

$\therefore$ $ T=83.54\,minutes $

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

If the length of a clock pendulum increase by $0.2\%$ due to atmospheric temperature rise, then the loss in time of clock per day is 

  1. $86.4$s
  2. $43.2$s
  3. $72.5$s
  4. $32.5$s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$T=2\pi \sqrt{\dfrac{l}{g}}$

$\dfrac{\Delta T}{T}\times 100 = \dfrac{1}{2}\dfrac{\Delta l}{l}\times 100 $

$\dfrac{\Delta T}{24 \times 3600}\times 100 = \dfrac{1}{2}\dfrac{0.2}{100}\times 100 $

$\Delta T=86.4s$