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Oscillations Test - 19

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Oscillations Test - 19
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  • Question 1
    1 / -0
    A force $$F=-10x+2$$ acts on a particle of mass $$0.1$$kg, where '$$k$$' is in m and $$F$$ in newton.If it is released from rest at $$x=-2$$m,find:(a) amplitude; (b) time period; (c) equation of motion.
    Solution
    Force, $$F=-10x+2$$
    $$F=-10(x-\dfrac{1}{5})$$
    Let $$y=x-\dfrac{1}{5}$$, $$F=-10y$$
    Now $$F=-kx$$
    So, $$k=10N/m$$

    a) Amplitude = Maximum displacement $$=|y| =|x-\dfrac{1}{5}| = |(-2)-\dfrac{1}{5}| = \dfrac{11}{5}$$ 
     
    b) Now, $$\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{10}{0.1}}=10 rad/s$$
    Time period, $$T=\dfrac{2\pi}{\omega} = \dfrac{2\pi}{10} =\dfrac{\pi}{5} sec$$

    c)  Finally, equation of motion will be given by: 
    $$y=Acos(\omega t+\phi)$$ 
    Now at $$t=0$$ when displacement is maximum. 
    $$-\dfrac{11}{5} = \dfrac{11}{5}cos\phi $$
    So, $$\phi = cos^{-1}(-1) =  \pi$$
    Therefore, $$y=Acos(\omega t +\pi) = -Acos(\omega t)$$ 
    Substitute for y in terms of x
    $$x-\dfrac{1}{5}=-\dfrac{11}{5}cos(\omega t)$$
    $$x=\dfrac{1}{5} - \dfrac{11}{5}cos(\omega t)$$
  • Question 2
    1 / -0
    A particle executes SHM with time period $$T$$ and amplitude $$A$$.The maximum possible average velocity in time $$\displaystyle{\dfrac{T}{4}}$$ is
    Solution
    Maximum average velocity magnitude for $$\dfrac{T}{4}$$ time will occur in the time from $$\dfrac{-T}{8}$$  to $$0$$ to $$\dfrac{+T}{8}$$,
    we can Integrate $$A \omega \cos{\omega t} {dt}$$ within the limits from $$\dfrac{-T}{8}$$ to $$\dfrac{+T}{8}$$, and divide the answer by $$\dfrac{T}{4}$$
    here $$\omega=\dfrac{2\pi}{T}$$, that gives, the maximum average velocity $$4\sqrt{2} \dfrac{A}{T}$$.
  • Question 3
    1 / -0
    The dimensional formula for amplitude of SHM is
    Solution
    Amplitude is the maximum displacement from mean level. Thus it has dimensions of Length or $$M^0LT^0$$
  • Question 4
    1 / -0
    What is the number of degrees of freedom of an oscillating simple pendulum?
    Solution
    There are 2 degrees of freedom, 1 translational along which the pendulum bob moves, and second rotational - the hinge about which it forms an arc motion.
  • Question 5
    1 / -0
    A particle is moving on a circle with uniform speed. Its motion is
    Solution
    Since the motion repeats itself periodically due to constant speed, thus it is periodic motion. But it is not to and fro motion (or SHM) because the acceleration is not dependent on displacement.
  • Question 6
    1 / -0
    Two particles P and Q describe SHM of same amplitude a and frequency v along the same straight line. The maximum distance between the two particles is $$a\sqrt{2}$$. The initial phase difference between the particle is
    Solution
    Let initial phase difference be $$\theta$$
    So, distance between two particales is $$d=asin(\omega t+\theta)-asin(\omega t)=2asin(\theta/2)cos(\omega t+\theta/2)$$
    As maximum value of $$d$$ is $$a\sqrt 2$$, so $$2asin(\theta/2)=a\sqrt 2\Rightarrow \theta=\pi/2$$
  • Question 7
    1 / -0
    The phase difference between the velocity and displacement of a particle executing SHM is
    Solution
    $$x=Asin(\omega t +\theta)$$
    $$v=dx/dt=A\omega cos(\omega t + \theta)=-A\omega sin (\omega t + \theta + \pi/2)$$ since $$cos(\pi/2+x)=-sin x$$ 
    Thus, phase difference is of $$\pi/2$$
  • Question 8
    1 / -0
    The displacement of a particle executing SHM at any instant t is $$x=0.01 \sin{100(t+0.05)}$$ then its time period will be
    Solution
    $$x=0.01\sin(100t+5)$$
    We have $$x=A\sin(\omega t + \theta)$$
    On comparing we get
    Thus, here $$\omega=100$$
    Thus, $$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{100}=0.06\ \text{seconds}.$$
  • Question 9
    1 / -0
    Which of the following characteristics must remain constant for undamped oscillations of the particle?
    Solution
    Acceleration is proportional to displacement from mean, thus it changes.
    Velocity also changes as it is zero at extreme and maximum at centre. 
    Phase also changes, but for the oscillations to be undamped, it must go to the same extreme position each time, or amplitude must be constant
  • Question 10
    1 / -0
    A person wearing a wrist watch that keeps correct time at the equator goes to N-pole. His watch will
    Solution
    Wrist watches don't work on the principle of gravity (like pendulum clocks). Thus there will be no effect.
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