Tag: a few applications of linear shm

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Questions Related to a few applications of linear shm

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

In a 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.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In simple harmonic motion, the average kinetic energy and the average potential energy over one complete period are equal, both being half of the total energy.

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

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:

  1. $\sqrt{2}\pi$
  2. $16\sqrt{2} \pi$
  3. $ \dfrac{32\sqrt{2}}{\pi}$
  4. $ \dfrac{4}{\pi}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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 :

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

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

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 :

  1. 3.0

  2. 2.0

  3. 1.5

  4. 1.0
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

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. $[K = mw^{2}]$.

  1. $0.02\ N/m$.
  2. $0.032\ N/m$.
  3. $0.132\ N/m$.
  4. $0.232\ N/m$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is d^2x/dt^2 + 16x = 0. Comparing this to d^2x/dt^2 + w^2x = 0, we get w^2 = 16, so w = 4 rad/s. Given mass m = 2g = 0.002 kg, the force constant K = m * w^2 = 0.002 * 16 = 0.032 N/m.

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

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

  1. 4 sec

  2. 2 sec

  3. 6 sec

  4. 3 sec
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $0\,to\,\,6.5$
$\begin{array}{l} 6.5=13\sin  \left( { \frac { { 2\pi t } }{ { 12 } }  } \right)  \ 0.5=\sin  \left[ { \left( { \frac { \pi  }{ 6 }  } \right) t } \right]  \ t=1\, \sec   \ total\, \, time=\, 2\times 1=2 \end{array}$

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

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

  1. simple harmonic motion

  2. non-periodic motion

  3. periodic but not simple harmonic motion

  4. oscillatory but not simple harmonic motion
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  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 $\pi/2$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

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

  1. 1/4 sec

  2. 1/2 sec

  3. 2sec

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

For a physical pendulum (ring oscillating about a nail on its circumference), the time period T = 2*pi * sqrt(I / mgd). Here, I = I_cm + md^2 = (1/2)mr^2 + mr^2 = (3/2)mr^2. With d = r, T = 2*pi * sqrt((3/2)mr^2 / mgr) = 2*pi * sqrt(3r / 2g). With diameter 1m, r = 0.5m. T = 2*pi * sqrt(1.5 / 9.8) approx 2 seconds.