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Oscillations Test - 9

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Oscillations Test - 9
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  • Question 1
    1 / -0
    A mass $$m$$ is undergoing SHM in the vertical direction about the mean position $$y_{0}$$ with amplitude A and angular frequency $$\omega$$. At a distance $$y$$ from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of $$m$$. Find the distance $$y^{*}$$ (measured from the mean position) such that the height $$h$$ attained by the block is maximum. $$(\mathrm{A}\mathrm{\omega})^{2}>g$$

    Solution
    Let   $$v$$  be the upward velocity of the block when it detaches from the spring at a distance $$y$$ above the mean position.
    $$v = w  \sqrt{A^2 - y^2}              \implies    v^2  = w^2   (A^2  - y^2)$$       ..........(1)
    At the detaching point, the block will continue to move upwards due to inertia and reach the  height  $$h$$ above the mean position. At that height the velocity of the block will be zero.
    Thus        $$0 - v^2  =  2(-g)s               \implies   s = \dfrac{v^2}{2g}$$
    Now the total height     $$h =s + y  =  \dfrac{v^2}{2g} + y $$
    $$ \therefore   h = \dfrac{w^2  (A^2 - y^2)}{2g}  + y $$
    For  $$h$$  to be maximum,     $$\dfrac{dh}{dy} = 0$$
    Thus       $$\dfrac{w^2}{2g}   (-2    y)   +  1   = 0$$
    $$\implies     y  =  \dfrac{g}{w^2}$$

  • Question 2
    1 / -0
    A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance $$\displaystyle\frac{2A}{3}$$ from equilibrium position. The new amplitude of the motion is.
    Solution
    In SHM, $$V=\sqrt { { A }^{ 2 }-{ x }^{ 2 } } $$

    $$V=\sqrt { { A }^{ 2 }-\dfrac { 4{ A }^{ 2 } }{ 9 }  } =\sqrt { \dfrac { 5{ A }^{ 2 } }{ 9 }  } $$

    $$3V=\sqrt { { A' }^{ 2 }-\dfrac { 4{ A }^{ 2 } }{ 9 }  } $$

    Dividing the above two equations give $$A'=\dfrac { 7A }{ 3 } $$
  • Question 3
    1 / -0
    A small block is connected to one end of a massless spring of unstretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t=0. lt then executes simple harmonic motion with angular frequency $$\omega =\frac { \pi }{ 3 } rad/s$$.
    Simultaneously at t=0, a small pebble is projected with speed v from point P at an angle of 45 as shown in the figure. Point P is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is
    (take $$g = 10 m/{ s }^{ 2 }$$)

    Solution
    Time is given by $$\displaystyle t=\frac{2vsin \theta}{g}$$
    $$\displaystyle 1= \frac{2vsin45^{o}}{10}$$
    $$\displaystyle v= \sqrt{50}m/s$$
  • Question 4
    1 / -0
    A particle is executing SHM along a straight line. Its velocities at distances $$x_1$$ and $$x_2$$ from the mean position are $$V_1$$ and $$V_2$$, respectively. Its time period is:
    Solution

    Using the expressions for velocity for SHM,

    $$v_1 = \omega \sqrt{A^2 -x_1^2}$$
    $$v_2 = \omega \sqrt{A^2 -x_2^2}$$
    Squaring both,
    $$v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)$$ ....(3)
    $$v_2 ^2 = \omega ^2 ( A^2 -x_2 ^2)$$ ....(4)
    Subtracting (4) from (3), $$v_1 ^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2) $$
    $$ \Rightarrow  4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 )$$
    $$ \Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2 ^2}}$$ where we have used $$\omega = \dfrac{2\pi}{T}$$

  • Question 5
    1 / -0
    Average velocity of a particle executing SHM in one complete vibration is : 
    Solution
    In one complete vibration, displacement is zero. So, average velocity in one complete vibration 
    $$= \dfrac{Displacement}{Time \, interval} = \dfrac{y_f - y_i}{T} = 0$$
    Hence option D is correct answer
  • Question 6
    1 / -0
    A simple pendulum performs simple harmonic motion about $$x = 0$$ with an amplitude $$'a'$$ and time period $$'T'$$. The speed of the pendulum at $$x = \dfrac{a}{2}$$ will be:
    Solution
    As simple pendulum performs simple harmonic motion.
    $$\therefore velocity, v = \omega \sqrt {a^{2} - x^{2}}$$
    At, $$x = \dfrac {a}{2}$$
    $$v = \dfrac {2\pi}{T} \sqrt {a^{2} -\left (\dfrac {a}{2}\right )^{2}} = \dfrac {2\pi}{T} \dfrac {\sqrt {3a^{2}}}{2} = \dfrac {\pi a\sqrt {3}}{T}$$
  • Question 7
    1 / -0
    A particle is describing simple harmonic motion. If its velocities are $$v_{1}$$ and $$v_{2}$$ when the displacements from the mean position are $$y_{1}$$ and $$y_{2}$$ respectively, then its time period is
    Solution
    In simple harmonic motion,
    velocity $$v = \omega \sqrt {A^{2} - y^{2}}$$
    $$\therefore v_{1} = \omega \sqrt {A^{2} - y_{1}^{2}} \Rightarrow v_{1}^{2} = \omega^{2}A^{2} - \omega^{2}y_{1}^{2} .... (i)$$
    and $$v_{2} = \omega \sqrt {A^{2} - y_{2}^{2}} \Rightarrow v_{2}^{2} = \omega^{2}A^{2} -\omega^{2} y_{2}^{2} ..... (ii)$$
    Solving equations (i) and (ii), we get
    $$v_{2}^{2} - v_{1}^{2} = \omega^{2} (y_{1}^{2} - y_{2}^{2})$$
    $$\omega = \sqrt {\dfrac {v_{2}^{2} - v_{1}^{2}}{y_{1}^{2} - y_{2}^{2}}}$$
    $$\Rightarrow T = \dfrac {2\pi}{\omega} = 2\pi \sqrt {\dfrac {y_{1}^{2} - y_{2}^{2}}{v_{1}^{2} - v_{2}^{2}}}$$.
  • Question 8
    1 / -0
    If maximum speed of a particular in SHM is given by $$\displaystyle { V }_{ m }$$. What is its average speed?
    Solution
    Let the SHM be represented as $$x=A sin(\omega t+\phi)$$
    Thus the speed of a particle=$$\dfrac{dx}{dt}=\left|A\omega cos(\omega t+\phi)\right|$$
    $$\Rightarrow V_m=A\omega$$
    The average speed of the particle=$$\dfrac{\int_0^{(2\pi/\omega)}\left|A\omega cos(\omega t+\phi)\right|}{2\pi/\omega}$$
    $$=\dfrac{2A\omega}{\pi}$$
    $$=\dfrac{2}{\pi}V_m$$
  • Question 9
    1 / -0
    The velocity vector v and displacement vector x of a particle executing SHM are related as $$\dfrac{vdv}{dx}=-\omega^2x$$ with the initial condition $$V=v_0$$ at $$x=0$$. The velocity v, when displacement is x, is?
    Solution
    \begin{array}{l}We\;have\;\\\dfrac{vdv}{dx}=-\omega^2x\;\\or\\\int_{v_0}^vvdv=-\omega^2\int_0^xxdx\\\dfrac{v^2}2-\dfrac{v_0^2}2=-\dfrac{\omega^2x^2}2\\v=\sqrt{v_0^2-\omega^2x^2}\end{array} 
  • Question 10
    1 / -0
    In SHM the net force towards mean position is related to its displacement (x) from mean position by the relation
    Solution
    Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
    $$F=-kx$$
    $$\implies F\propto x$$
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