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Oscillations Test - 12

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Oscillations Test - 12
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Weekly Quiz Competition
  • Question 1
    1 / -0
    If a particle is oscillating on the same horizontal plane on the ground.
    Solution
    A particle oscillating on ground possess both kinetic as well as potential energy.
  • Question 2
    1 / -0
    What is the effect on the time period of a simple pendulum if the mass of the bob is doubled?
    Solution
    As $$T=2 \pi \sqrt{\dfrac{L}{g}}$$
    It can be seen that T is independent of mass.
  • Question 3
    1 / -0
    The function $$sin ^2(\omega t)$$ represents :
    Solution

  • Question 4
    1 / -0
    The equation of motion of a particle is $$x = a cos(\alpha t)^2$$. The motion is
    Solution
    The motion is oscillatory but not periodic because the oscillations are followed from the maximum displacement (negative extreme) at t=0, the corresponding echo is equal to π/2

  • Question 5
    1 / -0

    Directions For Questions


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    The equation of simple harmonic wave is given by $$y=5$$sin $$\dfrac{5}{2}(100 t - x)$$, where $$x$$ and $$y$$ are in meter and time is in second. The time period of the wave (in second) will be.
    Solution
    $$y =5$$sin $$\dfrac{\pi}{2}(100 t - x)$$
    $$y= A$$sin$$(\omega t - kx)$$
    $$\omega = \dfrac{2\pi}{T} $$
    $$T=\dfrac{2 \pi}{100}$$
    $$\Rightarrow T=0.06$$
    Hence, the correct option is $$(B)$$
  • Question 6
    1 / -0
    The displacement of a particle from its mean position (in meter) is given by $$Y= 0.2\ sin (10 \pi t+1.5 )\ cos\ (10 \pi t+1.5 )$$. 
    The motion of the particle is
    Solution

  • Question 7
    1 / -0
    Equation of SHM is x = 10 sin 10 $$\pi$$t. Then the time period is, x is in cm and t is in sec
    Solution
    Given,
    $$x=10\sin(10\pi t)$$

    Comparing with $$x=A\sin(\omega t)$$

    $$\omega=10\pi$$

    $$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{10\pi}=0.2$$

    Option $$\textbf B$$ is the correct answer
  • Question 8
    1 / -0
    Time period will be minimum when d is equal to
  • Question 9
    1 / -0
    When a spring having spring constant $$2Nm^{-1}$$ is stretched by 5 cm, the energy stored in it is: 
    Solution
    Correct answer should be B.

  • Question 10
    1 / -0
    For a body in $$S.H.M$$ the velocity is given by the relation $$v=\sqrt{144-16x^2}m/sec$$. The maximum acceleration is  
    Solution
    velocity of the particle in SHM, 
    $$v=\sqrt{144-16x^{2}}=4\sqrt{9-x^{2}}$$ 

    comparing with $$v=\sqrt{A^{2}-x^{2}}$$
     we get $$A=3m$$ $$w=4rad/s$$

     so, maximum acceleration, 
    $$a_{max}=Aw^{2}$$ 
              $$=3\times 4^{2}$$ 
    $$\Rightarrow a_{max}=48m/s^{2}$$
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