Tag: simple pendulum

Questions Related to simple pendulum

In a simple harmonic motion

physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. the potential energy is always equal to the kinetic energy

  2. the potential energy is never equal to the kinetic energy

  3. the average potential energy in any time interval is equal to the average kinetic energy in that time interval

  4. the average potential energy in one time period is equal to the average kinetic energy in this period.

Show answer
Correct Option: D

The force which tries to bring a body back to its mean position is called :

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. deforming force

  2. restoring force

  3. gravitational force

  4. buoyant force

Show answer
Correct Option: B
Explanation:

At the extreme position, kinetic energy is zero and we know to throw out S.H.M.

K.E+P.E = constant
So, P.E. should be maximum at the extreme position.

A particle executes $SHM$ with a time period of $16\ s$. At time $t=2\ s$, the particle crosses the mean position while at $t=4s$, its velocity is $4ms^{-1}$. The amplitude of motion in meter is:

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $\sqrt{2}\pi$

  2. $16\sqrt{2} \pi$

  3. $ \dfrac{32\sqrt{2}}{\pi}$

  4. $ \dfrac{4}{\pi}$

Show answer
Correct Option: C
Explanation:

Let the equation of $S.H.M$ is:-


$x=a\sin\left(\dfrac{2\pi }{T}t+\phi\right)$

when $t=2s, x=0$ and $T=16s$ So,

$0=a\sin \left(\dfrac{\pi}{4}+\phi\right)$

Or $\phi=-\dfrac{\pi}{4}$

Therefore the eqn of $S.H.M$ is:-

$x=a\sin =\left(\dfrac{2\pi}{T}t-\dfrac{\pi}{4}\right)$

Now at time $t=4s, V=4m/s$

 So
$V=d\times dt=a\times \dfrac{2\pi}{T}\cos\left(\dfrac{2\pi}{T}t-\dfrac{\pi}{4}\right)$

So, $4=a\times \dfrac{2\pi}{16}\cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)$

Or, $4=a\times \dfrac{\pi}{8}\times \dfrac{1}{\sqrt{2}}$

Or $a=\dfrac{32\sqrt{2}}{\pi}$

Hence option $C$ is correct

The particle is executing S.H.M. on a line 4 cms long. If its velocity at its mean position is 12 cm/sec, its frequency in Hertz will be :

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $\dfrac{2\pi}{3}$

  2. $\dfrac{3}{2\pi}$

  3. $\dfrac{\pi}{3}$

  4. $\dfrac{3}{\pi}$

Show answer
Correct Option: B
Explanation:

Given,


$A=4cm$


$v=12cm/s$ at $x=0$ mean position

The velocity of particle performing S.H.M is given by

$v=\omega \sqrt{A^2-x^2}$

$12=\omega \sqrt{4^2-0}$

$12=4\omega$

$\omega =2\pi f=3$

$f=\dfrac{3}{2\pi}$

The correct option is B.

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15 ${ cms }^{ -1 }$ and the period is 628 milli-seconds. The amplitude of the motion in centimeters is :

physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. 3.0

  2. 2.0

  3. 1.5

  4. 1.0

Show answer
Correct Option: C
Explanation:

Given,


$T=628ms=0.628s$


$v _{max}=15cm/s=0.15m/s$

The maximum speed of the object is given by

$v _{max}=A\omega=A\dfrac{2\pi}{T}$

Amplitude, $A=\dfrac{v _{max}T}{2\pi}$

$A=\dfrac{0.15\times 0.628}{2\times 3.14}=0.015 m$

$A=1.5cm$

The correct option is C.

The different equation for linear SHM of a partial of mass $2g$ is $\dfrac {d^{2}x}{dt^{2}} + 16x = 0$. Find the force constant. $[K = mw^{2}]$.

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $0.02\ N/m$.

  2. $0.032\ N/m$.

  3. $0.132\ N/m$.

  4. $0.232\ N/m$.

Show answer
Correct Option: B

If a body mass $36 gm$ moves with S,H,M of amplitude $A=13$ and period  $T=12 sec$. At a time $t=0$ the displacement is $x=+13 cm$. The shortest time of passage from $x=+6.5$ cm to $x=-6.5$ is

physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. 4 sec

  2. 2 sec

  3. 6 sec

  4. 3 sec

Show answer
Correct Option: B
Explanation:

$\begin{array}{l} m=3bg,\, A=13,T=125 \ displacementx\left( t \right) =13\sin  \left( { \frac { { 2\pi t } }{ T }  } \right)  \end{array}$

Shortest time is at maximum slope which crosses zero. It will be from $ - 6.5\,\,to\,\,6.5\,\,$ or 2 times from $0\,to\,\,6.5$
$\begin{array}{l} 6.5=13\sin  \left( { \frac { { 2\pi t } }{ { 12 } }  } \right)  \ 0.5=\sin  \left[ { \left( { \frac { \pi  }{ 6 }  } \right) t } \right]  \ t=1\, \sec   \ total\, \, time=\, 2\times 1=2 \end{array}$

A function of time given by $\left(\sin{\omega t}-\cos{\omega t}\right)$ represents

physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. simple harmonic motion

  2. non-periodic motion

  3. periodic but not simple harmonic motion

  4. oscillatory but not simple harmonic motion

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} \sin  \omega t-\cos  \omega t \ =\sqrt { 2 } \left[ { \frac { 1 }{ { \sqrt { 2 }  } } \sin  \omega t-\frac { 1 }{ { \sqrt { 2 }  } } \cos  \omega t } \right]  \ =\sqrt { 2 } \left[ { \sin  \omega t\times \cos  \frac { \pi  }{ 4 } -\cos  \omega t\times \sin  \frac { \pi  }{ 4 }  } \right]  \ =\sqrt { 2 } \sin  \left( { \omega t-\frac { \pi  }{ 4 }  } \right)  \ this\, \, function\, \, represents\, \, SHM\, \, as\, \, it\, \, can\, \, be\, \, written\, \, in\, \, the\, \, form: \ a\sin  \left( { \omega t+\phi  } \right)  \ its\, \, period\, \, is,\, \, \frac { { 2\pi  } }{ \omega  }  \end{array}$

Hence,
option $(A)$ is correct answer.

A particle is subjected to two simple harmonic motions along $x$ and $y$ directions according to $x=3\sin\ 100\pi t$ $y=4\sin\ 100\pi t$

physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. Motion of particle will be on ellipse travelling in clockwise direction.

  2. Motion of particle will be on a straight line with slope $4/3$

  3. Motion will be simple harmonic motion with amplitude $5$.

  4. Phase difference between two motions is $\pi/2$.

Show answer
Correct Option: A

A ring whose diameter is 1 meter, oscillates simple harmonically in a vertical plane about a nail fixed at its circumference and perpendicular to plane of ring. The time period will be

physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. 1/4 sec

  2. 1/2 sec

  3. 2sec

  4. None of these

Show answer
Correct Option: C