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Oscillations Test - 21

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Oscillations Test - 21
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  • Question 1
    1 / -0
    A body of mass 100 gm is suspended from a spring of force constant 50 N/m. The maximum acceleration produced in the spring is:
    Solution
    To calculate maximum acceleration consider spring at the equilibrium position: 
    $$F=ma$$
    where $$F$$ is force on spring which will be equal to $$mg$$ only (as gravity is the only force present. 
    Therefore, $$ma=mg$$ or $$a=g$$.


  • Question 2
    1 / -0
    Restoring force in the SHM is
    Solution
    A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. Restoring force $$F=kx,$$ is such kind of force and is conservative.
  • Question 3
    1 / -0
    Due to some force $$F_1$$, a body oscillates with period 4/5 second and due to other force $$F_2$$  it oscillates with the period of 3/5 sec. If both forces act simultaneously new period will be
    Solution
    for first case, $$\omega_1^2=\dfrac{F_1}{mx}$$
    for second case, $$\omega_2^2=\dfrac{F_2}{mx}$$
    When both forces applied, $$\omega_3^2=\dfrac{F_1+F_2}{mx}$$
    $$\Rightarrow \omega_3^2= \omega_1^2+ \omega_2^2$$
    $$\Rightarrow \omega_3^2/4\pi^2= \omega_1^2/4\pi^2+ \omega_2^2/4\pi^2$$
    $$\dfrac{1}{T_3^2}=\dfrac{1}{T_1^2}+\dfrac{1}{T_2^2}$$
    $$T_1=4/5=0.8s;T_2=3/5=0.6s$$
    $$\Rightarrow T_3=12/25=0.48\ sec$$
  • Question 4
    1 / -0
    A particle executes simple harmonic motion with frequency 2.5 Hz and amplitude 2 m. The speed of the particle 0.3 s after crossing the equilibrium position is:
    Solution
    $$\displaystyle T=\frac{1}{F}=\frac{1}{2.5}=0.4$$ s
    $$\displaystyle t=0.3s=\frac{3T}{4}$$
    At the given time, particle win be in its extreme position if at t=0 it crosses the mean position.
  • Question 5
    1 / -0
    Statement 1: The graph between velocity and displacement for a harmonic oscillator is an ellipse.

    Statement 2: Velocity does not change uniformly with displacement in harmonic motion.
    Solution
    $$x=Asin\omega t$$ or $$sin\omega t=x/A$$
    $$v=dx/dt=A\omega cos\omega t$$ or $$cos\omega t=v/A\omega$$
    Thus, using $$sin^2+cos^2=1$$, $$x^2/A^2+x^2/A^2\omega^2=1$$, which gives the equation of ellipse.
    Thus statement 1 is correct. Also, statement 2 is correct but it doesn't explain the reason.
  • Question 6
    1 / -0
    A mass is suspended separately by two springs of spring constant $$k_{1}$$ and $$k_{2}$$ in successive order. The time periods of oscillations in the two cases are $$T_{1}$$ and $$T_{2}$$ respectively. If the same mass be suspended by connecting the two springs in parallel $$($$as shown in figure$$)$$ then the time period of oscillations is $$T.$$ The correct relation is

    Solution
    $$\displaystyle T=2\pi \sqrt{\frac{M}{k}}$$
    $$\therefore $$   $$\displaystyle k=\frac{4\pi ^{2}m}{T^{2}}$$
    In the given situation
    $$k=k_{1}+k_{2}$$
    $$\therefore $$   $$\displaystyle \frac{4\pi ^{2}m}{T^{2}}=\frac{4\pi ^{2}m}{T_{1}^{2}}+\frac{4\pi ^{2}m}{T_{2}^{2}}$$
    $$\therefore $$   $$T^{-2}=T_{1}^{-2}+T_{2}^{-2}$$
  • Question 7
    1 / -0
    A simple harmonic oscillator has an acceleration of $$1.25\ m/s^{2}$$ at $$5\ cm$$ from the equilibrium. Its period of oscillation is:
    Solution
    $$\displaystyle T=2\pi \sqrt{\left | \frac{x}{a} \right |}=2\pi \sqrt{\frac{5\times 10^{-2}}{1.25}}=\frac{2\pi }{5}$$ sec
  • Question 8
    1 / -0
    A particle moves in a circular path with a uniform speed. Its motion is:
    Solution

    $$\textbf{Correct option: Option (A).}$$

    $$\textbf{Explanation:}$$

    The motion of particle moving in a circular path with uniform speed is periodic and is not oscillatory because it’s not to and fro. Hence it can also not be SHM and angular SHM.

    $$\textbf{Thus, option (A) is only correct.}$$

  • Question 9
    1 / -0
    Statement 1: In simple harmonic motion, the motion is to and fro and periodic.

    Statement 2: Velocity of the particle $$(v) = \omega\sqrt{k^2 - x^2}$$ (where $$x$$ is the displacement).
    Solution
    $$S.H.M$$ is to and fro motion of an object and it is periodic. 
    $$v=\omega \sqrt{k^2 - x^2}$$
    If $$x = 0$$, $$v$$ has maximum value. At $$x = k$$, $$v$$ has minimum velocity. Similarly, when $$x = -k$$, $$v$$ has zero value, all these indicate to and fro movement.
  • Question 10
    1 / -0
    A particle of mass 2 kg moves in simple harmonic motion and its potential energy U varies with position x as shown. The period of oscillation of the particle is:

    Solution
    $$\displaystyle \frac{1}{2}kA^{2}=1.0$$
    $$\displaystyle \frac{1}{2}k\left ( 0.4 \right )^{2}=1.0$$
    or $$k=12.5$$ or $$\displaystyle \frac{25}{2}$$ N/m
    $$\displaystyle T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{2}{\frac{25}{2}}}=\frac{4\pi }{5}$$ s
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