Tag: oscillatory motion

Questions Related to oscillatory motion

A pendulum bob has a speed of $ 3 $ $ \mathrm{ms}^{-1} $ at  its lowest position. The pendulum is$ 0.5$ $ \mathrm{m}  $ long. The speed of the bob, when the length makes an angle of $ 60^{\circ}  $ to the vertical will be $ (g=10 $ $ \left(n s^{-1}\right) $

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

  1. $3$ $ m s^{-1} $

  2. $
    1 / 3 \mathrm{ms}^{-1}
    $

  3. $
    1 / 2 m s^{-1}
    $

  4. $
    2 m s^{-1}
    $

Show answer
Correct Option: D
Explanation:
Apply energy conservation theorem$,$ 
energy at lowest position of Bob $=$ energy$ ,$ when Bob makes $60°$ to the vertical 
$1/2 mv^2 = 1/2 mv₁^2 + mgl(1 - cos60°)$
Here $v$ is speed at Lowest position $, v₁$ is speed $,$ when it makes $60°$ with vertical and $l$ is length of pendulum $.$
$[$Actually, height of Bob $,$ when it makes $60°$ with vertical $= l(1 - cos60°)] $
$∴ v^2 = v₁^2 + 2gl(1 - cos60°)$ 
$3^2 = v₁^2 + 2 × 10 × 0.5 (1 - 1/2)$ 
$9 = v₁^2 + 5$ 
$v₁^2 = 4 ⇒v₁ = 2m/s $
So$,$ speed of Bob $= 2m/s$
Hence,
option $(D)$ is correct answer.

Which of the following will change the time period as they are taken to moon?

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

  1. A simple pendulum

  2. A physical pendulum

  3. A torsional pendulum

  4. A spring-mass system

Show answer
Correct Option: A,B
Explanation:

$(i)$ For simple pendulum $T = 2\pi\sqrt{L/g}$
$(ii)$ For physical pendulum $T = 2\pi\sqrt{I/mgL}$
So in both above case, time period is changed if they are taken to the moon.
$(iii)$ For torsional pendulum $T = 2\pi\sqrt{I/C}$
$(iv)$ For spring-mass system $T = 2\pi\sqrt{m/k}$

A simple pendulum of length L and having a bob of mass m is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium positions, its time period of oscillation is:

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

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

  2. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}+ \dfrac{v^4}{R^2}}}$

  3. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}+ \dfrac{v^2}{R}}}$

  4. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}- \dfrac{v^4}{R^2}}}$

Show answer
Correct Option: B
Explanation:

The period of the pendulum, $T=2\pi \sqrt{\cfrac{L}{a}}$ where $a=$ $\text{resultant acceleration}\=\sqrt{g^2+{(\cfrac{V^2}{R})}^2}\quad\quad\quad\quad [\cfrac{V^2}{R}=\text{centripital acceleratiop }, g=\text{acceleration due to gravity}]$.

$\therefore T=2\pi\sqrt{\cfrac{L}{g^2+\cfrac{V^4}{R^2}}}$
Option B is the correct answer.


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

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.

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

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Correct Option: C

Sitar maestro Ravi Shankar is playing sitar on its strings, and you, as a physicist (unfortunately without musical ears!), observed the following oddities.
I. The greater the length of a vibrating string, the smaller its frequency.
II. The greater the tension in the string, the greater is the frequency.
III. The heavier the mass of the string, the smaller the frequency.
IV. The thinner the wire, the higher its frequency.
The maestro signalled the following combination as correct one.

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

  1. II, III and IV

  2. I, II and IV

  3. I, II and III

  4. I, II, III and IV

Show answer
Correct Option: D
Explanation:

Guitar string is a standing wave as both the ends of guitar string are fixed.

$f \propto \cfrac{1}{length}$
$f \propto \sqrt{tension}$
$f \propto \cfrac{1}{\sqrt{\mu}}$, where $\mu$ is mass per unit length.
$f \propto \cfrac{1}{\text{thickness of wire}}$