Self Studies

Oscillations Test - 15

Result Self Studies

Oscillations Test - 15
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    A man of mass 60 kg, standing on a platform, is executing SHM in a vertical plane. The displacement from mean position is $$y = 0.5 sin (2\pi t)$$. The minimum value of frequency ($$f$$) for which the man will feel weightlessness at the highest point is :
    Solution
    $$w=2\pi

    f$$
    $$A=0.5$$
    $$a=-w^{2}A$$
    For weightlessness we have $$a=g$$
    Thus
    $$g=-w^{2}A$$
    $$w^{2}=\dfrac{g}{0.5}=2g$$
    $$4\pi^{2}f^{2}=2g$$
    $$f^{2}=\dfrac{g}{2\pi

    ^{2}};f=\dfrac{\sqrt{2g}}{2\pi}$$
  • Question 2
    1 / -0
    $$ T _{1}$$, $$ T _{2}$$ are time periods of oscillation of a block individually suspended to spring of force constants $$ K _{1}$$ ,$$ K _{2}$$ respectively. If same block is suspended to parallel combination of same two springs, its time period is
    Solution


    for a parallel combination $$k=k_{1}+k_{2}$$
    $$T=2\pi \sqrt{\dfrac{m}{k}}$$

    $$=2\pi \sqrt{\dfrac{m}{(k_{1}+k_{2})}}$$

    $$T_{1}=2\pi \sqrt{\dfrac{m}{k_{1}}}$$

    $$\Rightarrow k_{1}=(2\pi )^{2}\dfrac{m}{T_{1}^{2}}$$

    $$k_{2}=(2\pi )^{2}\dfrac{m}{T_{2}^{2}}$$

    $$k_{1}+k_{2}=(2\pi )^{2}(\dfrac{m}{T_{1}^{2}}+\dfrac{m}{T_{2}^{2}})$$

    $$\therefore \dfrac{T_{1}T_{2}}{\sqrt{T_{1}+T_{2}}}=T$$


  • Question 3
    1 / -0
    An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $$15 cm/sec$$ and the period is $$628$$ milliseconds. The amplitude of the motion in centimeters is:                           
    Solution

    $$v_{max}=15\ cm /s ; T=0.628\ s=\dfrac{2\pi}{10}s$$
    $$\omega =\dfrac{2\pi }{T}=10\ rad/s$$
    $$v_{max}=A\omega $$
    $$A=1.5\ cm$$

  • Question 4
    1 / -0
    The time period of oscillation of a particle that executes S.H.M. is $$1.2 sec$$. The time starting from extreme position, its velocity will be half of its velocity at mean position is 
    Solution
    Time is zero at extreme position
    $$T=1.2 sec $$
    $$v=\omega \sqrt{A^{2}-x^{2}} ; v_{max}=\omega A$$
    $$v=\dfrac{v_{max}}{2}=\dfrac{\omega A}{2}=\omega$$ $$\sqrt{A^{2}-x^{2}}$$
    $$\dfrac{A^{2}}{4}=A^{2}-x^{2} ; x=\dfrac{\sqrt{3}}{2}A$$
    $$x=A$$ cos $$\omega t$$
    $$\dfrac{\sqrt{3}}{2}A=A$$ cos $$\omega t$$
    $$\omega t=\dfrac{\pi }{6} $$

    $$ t=\dfrac{\pi }{6\times \omega }$$
    $$t=\dfrac{\not{\pi} }{6}\times \dfrac{1.2}{{2\not \pi }}=0.1 sec$$
  • Question 5
    1 / -0
    A simple harmonic oscillation is represented by the equation $$ y = 0.4 \sin(440t/7+0.61) $$ where $$y$$ and $$t$$ are in meters and seconds respectively, the value of time period is :
    Solution
    $$\omega =\dfrac{440}{7}$$

    $$\dfrac{2\pi }{T}=\dfrac{440}{7} ; T=\dfrac{14\pi }{440}$$

    $$T=0.1 sec$$
  • Question 6
    1 / -0
    The equation for the displacement of a particle executing SHM is y = {5 sin 2$$ \pi $$ t }cm. Then the velocity at 3 cm from the mean position is 
    Solution
    $$\omega =2\pi \ {rad}/{sec}$$
    $$A=5cm$$
    $$x=3cm$$
    $$v=\omega \sqrt{A^{2}-x^{2}}$$
    $$=2\pi \times \sqrt{25-9}=8\pi \ {cm}/{s}$$
  • Question 7
    1 / -0
    A particle executing SHM has velocities $$20$$ cm/s and $$16$$ cm/s at displacements $$4$$ cm and $$5$$ cm from its mean position respectively. Its time period is
    Solution
    $$v_{1}=\omega _{1}\sqrt{A^{2}-x_{1}^{2}}$$
    $$v_{2}=\omega _{1}\sqrt{A^{2}-x_{2}^{2}}$$
    $$\dfrac{v_{1}}{v_{2}}=\dfrac{\sqrt{A^{2}-x{_{1}}^{2}}}{\sqrt{A^{2}-x{_{2}}^{2}}}

    ;

    v{_{1}}^{2}A^{2}-v{_{1}}^{2}x{_{2}}^{2}=v{_{2}}^{2}A^{2}-v{_{2}}^{2}x{_{1}}^{2}$$
    $$A^{2}=\dfrac{v{_{1}}^{2}x{_{2}}^{2}-v{_{2}}^{2}x{_{1}}^{2}}{v{_{1}}^{2}-v{_{2}}^{2}}=\dfrac{400\times

    25-16\times 256}{400-256}$$
    $$A^{2}=41 $$
    Substituting  $$A^{2}=41

    $$ we get $$\omega ^{2}=16$$
    $$\omega =\dfrac{2 \pi}{T}=4 ; T=\dfrac{2\pi }{4}=\dfrac{\pi
    }{2}sec$$

  • Question 8
    1 / -0
    The velocity of a particle in SHM at the instant when it is $$0.6$$ cm away from the mean position is $$4$$ cm /sec. If the amplitude of vibration is $$1$$ cm then its velocity at the instant when it is $$0.8$$cm away from the mean position is:
    Solution

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

    $$V_{1}=\omega \sqrt{A^{2}-x^{2}}$$

    $$V_{1}=\omega \sqrt{A^{2}-(0.6)^{2}}$$ 

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

    $$V_{2}=\omega \sqrt{A^{2}-(0.8)^{2}}$$

    $$\dfrac{V_{1}}{V_{2}}=\dfrac{\sqrt{(1^{2}-(0.6)^{2}}}{\sqrt{(1^{2}-(0.8)^{2}}}$$

    $$\dfrac{4}{V_{2}}=\dfrac{0.8}{0.6}$$

    $$V_{2}=3  \ cm / sec$$

  • Question 9
    1 / -0
    Two S.H.M.'s are represented by the relations $$ x=4sin(80t+\pi /2) $$ and $$ y=2cos(60t+\pi/3) $$.
    The ratio of their time periods is
    Solution

    $$x=4 sin(80t +\dfrac{\pi }{2})$$
    $$\omega _{1}=80 ;\dfrac{2\pi}{T_{1}}=80 ; T_{1}=\dfrac{\pi

    }{40}$$sec
    $$y=2 cos(60t + \dfrac{\pi }{3})$$
    $$\omega _{2}=60 ; \dfrac{2\pi }{T_{2}}=60 ; T_{2}=\dfrac{\pi

    }{30}sec$$
    $$\dfrac{T_{1}}{T_{2}}=\dfrac{3}{4}$$

  • Question 10
    1 / -0
    The displacement of a particle executing S.H.M. is given by $$ y = 10 \sin (6t+\pi/3) $$ in metre and time in seconds. The initial displacement and velocity of the particle are respectively :
    Solution
    Initially $$t=0 ;y=10 sin(\dfrac{\pi }{3})$$

    $$=10\times \dfrac{\sqrt{3}}{2}=10\dfrac{\sqrt{3}}{2}m$$

    $$v=\dfrac{dy}{dt}=60 cos(6t + \dfrac{\pi }{3})$$

    At $$t=0 \Rightarrow 60\times cos \dfrac{\pi

    }{3}=30{m/}{s}$$
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now