Tag: example of simple harmonic motion

Questions Related to example of simple harmonic motion

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


Correct Option: B

A string of a simple pendulum can bear maximum tension that is $1.5$ times the weight of the bob. What is the maximum angle with which it can oscillate?

  1. $\cos^{-1} (0)$

  2. $\cos^{-1} (0.75)$

  3. $\cos^{-1} (0.25)$

  4. $\cos^{-1} (\sqrt {3}/2)$


Correct Option: A

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. $16:1$


Correct Option: C

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


Correct Option: C

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. $0.034\ N$


Correct Option: A

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

Correct Option: A
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 $

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


Correct Option: A
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 $

$\Delta T=86.4s$

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


Correct Option: B
Explanation:

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

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


Correct Option: A
Explanation:

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

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


Correct Option: A