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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 $

$\Delta T=86.4s$

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

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are
attached at distance $'L/2'$ from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio m/M is close to :

  1. 0.175

  2. 0.375

  3. 0.575

  4. 0.775
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency of torsonal oscillations is given by 
$f = \dfrac{k}{\sqrt{I}}$
$f _1 = \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12}}}$
$f _2 =  \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12} + 2m \left(\dfrac{L}{2} \right)^2}}$
$f _2 = 0.8 f _1$
$\dfrac{m}{M} = 0.375$

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

Find the time period of oscillations of a torsional pendulum, if the torsional constant of the wire is K = 10$\pi^2$J/rad. The moment of inertia of rigid body is 10 Kg m$^2$ about the axis of rotation.

  1. 2 sec

  2. 4 sec

  3. 16 sec

  4. 8 sec
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Time period of a torsional pendulum is given by
$T = 2 \pi \sqrt{\dfrac{I}{k}}$
$\Rightarrow T=2 \pi \sqrt{\dfrac{10}{10\pi^2}}=2 sec$

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

A clock which has a pendulum made of brass keep correct time at ${30^0}C$? How many seconds it will gain or lose in day if the temperature falls to ${0^0}C$.

  1. It will lose $23.32$ sec per day
  2. It will gain $23.32$ sec per day
  3. It will lose $50$ sec per day
  4. It will gain $50$ sec per day
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The change in time period is delta_T = (1/2)*alpha*T*delta_theta. For brass, alpha is approx 1.8e-5. delta_T/T = (1/2)*alpha*delta_theta. The time lost/gained per day is (delta_T/T) * 86400. Plugging in values gives approx 23.3 seconds.

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

A simple pendulum of length $1$m has a bob of mass $100$g. It is displaced through an angle of $60^o$ from the vertical and then released . Find out K.E. of bob when it passes through mean position.

  1. $0.12$J
  2. $0.24$J
  3. $0.36$J
  4. $0.55$J
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Length of simple pendulum $=1m$

Mass $=1w\;gm=0.1kg$
It is displaced through as angle of $60$ for vertical 
Height of pendulum at starting position $=$ length $(1-ws\;60)$
                                                                   $=1\left( 1-0.5\right)$
                                                                   $=0.5m$
Potential energy $=mgh=0.1\times 10\times 0.5 =0.55$
when it is released and it reaches mean position its potential energy at starting point is converted to kinetic energy.
so K.E. of bob at mean position $=0.55.$
Hence, the answer is $0.55.$

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

A pendulum is formed by pivoting a long thin rod of length L and mass m about a point P on the rod which is a distance d above the center of the rod as shown. 
Now answer the following questions. 
1. The time period of this pendulum when d = L/2 will be

  1. $2\pi\sqrt { \dfrac { 2\ell }{ 3g }}$
  2. $2\pi\sqrt { \dfrac { 3\ell }{ 2g }}$
  3. $4\pi\sqrt { \dfrac { \ell }{ 3g }}$
  4. $\dfrac {2\pi} {3} \sqrt { \dfrac { 2\ell }{ g }}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A uniform rope of mass M =0.1 kg and length L=10 m hangs from the ceiling. [$  g=10 m /{ s }^{ 2 } $]

  1. Speed of the transverse wave in the rope increases linearly from bottom to the top with distance.

  2. Speed of the transverse wave in the rope decreases linearly from bottom to the top with distance.

  3. Speed of the transverse wave in the rope remains constant along the length of the rope

  4. Time taken by the transverse wave to travel the full length of the rope is 2 sec
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The speed of a wave in a hanging rope is v = sqrt(T(x)/mu) = sqrt(mu*g*x/mu) = sqrt(gx). The time to travel the full length is integral(dx/sqrt(gx)) from 0 to L, which is 2*sqrt(L/g). With L=10 and g=10, time = 2*sqrt(1) = 2 seconds.

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

A simple pending of length l has a bob of mass m, with a charge q on it . A  vertical sheet of charge, with surface charge density $\sigma $ passes string makes an angle $\theta $ with the vertical , then 

  1. $\quad tan\theta =\dfrac { \sigma q }{ 2{ \epsilon } _{ 0 }mg } $
  2. $\quad tan\theta =\dfrac { \sigma q }{ { \epsilon } _{ 0 }mg } $
  3. $\quad cot\theta =\dfrac { \sigma q }{ 2{ \epsilon } _{ 0 }mg } $
  4. $\quad cot\theta =\dfrac { \sigma q }{ { \epsilon } _{ 0 }mg } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric field due to a sheet of charge is E = sigma / (2*epsilon_0). The force on the bob is qE. The angle theta satisfies tan(theta) = F_electric / F_gravity = (q*sigma / (2*epsilon_0)) / (mg).