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Oscillations Test - 30

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Oscillations Test - 30
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  • Question 1
    1 / -0
    An oscillator consists of a block attached to a spring (k = 400 N/m). At some time t, the position (measured from the system's equilibrium location), velocity and acceleration of the block are x = 0.100m, v = 13.6 m/s, and a = 123 m/s$$^2$$. The amplitude of the motion and the mass of the block are
    Solution

  • Question 2
    1 / -0
    A particle performs SHM along a straight line with the period $$T$$ and amplitude $$A$$. The mean velocity of the particle averaged over the time during which it travels a distance $$A/2$$ string from the extreme position is
    Solution
    The position of particle $$x=A\cos { \cfrac { 2\pi  }{ T } t } $$ (As the particle starting from mean position)
    Velocity $$v=-A\cfrac { 2\pi  }{ T } \sin { (\cfrac { 2\pi  }{ T } { t }) } $$
    Let it will take t' time in reaching from $$A$$ to $$\cfrac { A }{ 2 } $$. 
    So, $$\cfrac { A }{ 2 } =A\cos { (\cfrac { 2\pi  }{ T } { t' }) } $$
    $$\Rightarrow \cfrac { 2\pi  }{ T } { t' }=\cfrac { \pi  }{ 3 } $$
     $$\Rightarrow { t' }=\cfrac { T }{ 6 } $$
    So, the required mean velocity over this time period
    $${ v }_{ av }=\cfrac { \int _{ 0 }^{ \cfrac { T }{ 6 }  }{ v } dt }{ \int _{ 0 }^{ \cfrac { T }{ 6 }  }{ dt }  } $$
    $$=\cfrac { 6 }{ T } \int _{ 0 }^{ T }{ -A(\cfrac { 2\pi  }{ T } )\sin { (\cfrac { 2\pi  }{ T } t) } dt } $$
    $$=\cfrac { 6\times 2\pi A }{ { T }^{ 2 } } \times \cfrac { T }{ 2\pi  } [-\cos { (\cfrac { 2\pi  }{ T } t) } ]_{ 0 }^{ \cfrac { T }{ 6 }  }$$
    $$=\cfrac { 12\pi A }{ { T }^{ 2 } } [\cfrac { 1 }{ 2 } -1]\times \cfrac { T }{ 2\pi  } $$
    $$=-\cfrac { 3A }{ T } $$
    So, the magnitude of $${ v }_{ av }$$ is $$\cfrac { 3A }{ T } $$
  • Question 3
    1 / -0
    Three masses of $$500g,300g$$ and $$100g$$ are suspended at the end of a spring as shown and are in equilibrium. When the $$500g$$ mass is removed suddenly, the system oscillates with a period of $$2$$ second. When the $$300g$$ mass is also removed, it will oscillate with a period

    Solution
    After removing $$500g$$-
    $$T=2\pi\sqrt{\cfrac{m}{K}}=2sec\\ \quad=2\pi\sqrt{\cfrac{\cfrac{400}{100}}{K}}=2sec\rightarrow (1)$$
    Now after removing $$300$$g also
    $$\Rightarrow T=2pi\sqrt{\cfrac{\cfrac{100}{1000}}{K}}\rightarrow (2)$$
    From equation $$1$$ and $$2$$
    $$\cfrac{2}{T}=\cfrac{\sqrt{400}}{\sqrt{100}}\\ \cfrac{2}{T}=\cfrac{20}{10}\\T=1sec$$

  • Question 4
    1 / -0
    A particle performing S.H.M. having amplitude 'a' possesses velocity $$\dfrac {(3)^{1/2}}{2}$$ times the velocity at the mean position. The displacement of the particle shall be
    Solution
    In S.H.M the velocity of particle is 
    given by , $$\boxed{v^2=w^2(a^2-x^2)}$$
    Where, a-amplitude of SHM
                 x-displacement of particle
    At mean position $$(x=0)$$, let the velocity 
    of particle be $$v$$.
    $$\Rightarrow v^2=w^2a^2$$
    $$\Rightarrow \boxed{w^2=\dfrac{v^2}{a^2}}$$.....(1)
    Now, let at a displacement $$x$$, the velocity 
    of particle is $$\dfrac{\sqrt{3}}{2}v. $$  We have to find $$x$$.
    $$\left(\dfrac{\sqrt{3}}{2}v\right)^2=w^2\left(a^2-x^2\right)$$
    $$\Rightarrow (\dfrac{\sqrt{3}}{2}v)^2=\dfrac{v^2}{a^2}(a^2-x^2)$$     from (1)
    $$\Rightarrow \dfrac{3}{2}v^2=\dfrac{v^2}{a^2}(a^2-x^2)$$
    $$\Rightarrow 3a^2=4(a^2-x^2)$$
    $$\Rightarrow 3a^2=4a^2-4x^2$$
    $$\Rightarrow 4x^2=a^2\Rightarrow x^2=\dfrac{a^2}{4}$$
    $$\Rightarrow \boxed{x=\dfrac{a}{2}}$$
  • Question 5
    1 / -0
    The phase difference between the two simple harmonic oscillations, $$y_{1} = \dfrac {1}{2} \sin \omega t + \left (\dfrac {\sqrt {3}}{2}\right )\cos \omega t$$ and $$y_{2} = \sin \omega t + \cos \omega t$$ is
    Solution
    $$ y_1 = \dfrac{1}{2}\sin\omega t + \dfrac{\sqrt{3}}{2}\cos\omega t $$

      $$= \cos\left(\dfrac{\pi}{3}\right)\sin\omega t + \sin\left(\dfrac{\pi}{3}\right)\cos\omega t$$

     $$ = \sin\left(\dfrac{\pi}{3}+\omega t \right) $$



     $$y_2 = \sin\omega t + \cos\omega t$$

    $$  = \sqrt{2}\left(\dfrac{1}{\sqrt{2}}\sin\omega t +\dfrac{1}{\sqrt{2}}\cos\omega t\right)$$

    $$ =  \sqrt{2}\left(\cos\dfrac{\pi}{4}\sin\omega t + \sin\dfrac{\pi}{4}\cos\omega t\right) $$

    $$ = \sqrt{2}\sin\left(\dfrac{\pi}{4}+\omega t \right)$$

    Now phase difference 
     
    $$ \phi = \left(\dfrac{\pi}{3}+\omega t\right) - \left(\dfrac{\pi}{4}+\omega t\right)$$

      $$ = \dfrac{\pi}{12} $$
    Hence answer is option number- (C)
  • Question 6
    1 / -0
    The time period of simple harmonic motion depends upon the
    Solution
    The time period of simple harmonic motion depends upon the mass.
    For example:- Simple pendulum,
    The time period of the simple pendulum is given by
    $$T=2\pi \sqrt{\dfrac{m}{k}}$$. . . .. . . .(1)
    From the above equation we can conclude that,
    1) $$T\propto \sqrt{m}$$ (mass of the particle)
    2) $$T\propto \dfrac{1}{\sqrt{k}}$$ $$(k=force\,   constant)$$
    The correct option is D.
  • Question 7
    1 / -0
    An oscillator is producing FM waves of frequency 2kHz with avariation of 10kHz. The modulating index=?
    Solution
    Modulation Index , $$m_f=\dfrac{\text{Maximum frequency deviation}}{\text{Modulating frequency}}$$
    $$\Rightarrow m_f= \dfrac{10 kHz}{2 kHz}$$
    $$\Rightarrow m_f=5$$
  • Question 8
    1 / -0
    In an electronic watch, the component corresponding to the pendulum of a pendulum clock is a__?
    Solution

  • Question 9
    1 / -0
    A spring balance together with a suspended weight of $$2.5$$kg is dropped from a height of $$30$$ metres. The reading on the spring balance, while falling, will show a weight of.
    Solution
    Spring balance reads the net force acting on it by the suspended object. While free falling the object is accelerating with acceleration g m/s-2 . With respect to the spring a pseudo force acts on the object which is equal to mg opposite to the weight of the body. The pseudo force balances the weight of the body and the body does not exert any force on the spring. Thus the spring reading will be zero.
  • Question 10
    1 / -0
    In case of a simple harmonic motion, if the velocity is plotted along the X-axis and the displacement (from the equilibrium position)(is plotted along the Y-axis, the resultant curve happens to be an ellipse with the ratio:
    $$\displaystyle\frac{major axis(along X)}{minor axis (along Y)}=20\pi$$
    What is the frequency of the simple harmonic motion?
    Solution
    Relation between velocity & x is SHM in H.
    $$\displaystyle\frac{v^2}{\omega^2A^2}+\frac{x^2}{A^2}=1$$ $$\rightarrow$$ Ellipse
    Major axis $$=2\Omega A$$
    Minor axis $$=2a$$
    Given: $$\displaystyle\frac{2\Omega a}{2a}$$
    $$\Omega =20\pi$$
    $$2\pi f=20\pi$$
    $$f=10$$Hz.

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