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Oscillations Test - 22

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Oscillations Test - 22
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  • Question 1
    1 / -0
    Select the correct statements from the following
    Solution

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

    $$\textbf{Explanation:}$$

    (A) A simple harmonic motion is always repeated at a regular interval and hence it is necessarily periodic.

    (B) An SHM is a to and fro motion and hence it is necessarily oscillatory and hence this statement is incorrect.

    (C) An oscillatory motion need not be always periodic and hence this statement is incorrect.

    (D) A periodic motion may or may not be oscillatory. So this statement is also false.

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

  • Question 2
    1 / -0
    A particle performs SHM with a period $$T$$ and amplitude $$a$$. The mean velocity of particle over the time interval during which it travels a distance $$a/2$$ from the extreme position is:
    Solution
    Mean velocity$$\displaystyle =\frac{displacement}{time}$$
       $$\displaystyle =\frac{\left ( a/2 \right )}{\left ( T/6 \right )}=\frac{3a}{T}$$

  • Question 3
    1 / -0
    Find the length of a simple pendulum such that its time period is $$2\ s$$.
    Solution
    $$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$$

    $$T^2\, =\, 4\pi ^2\, \displaystyle \frac {L}{g}$$
    $$\Rightarrow\, 2^2\, =\, 4\, \times\, 3.14\, \times\, 3.14\, \times\, \displaystyle \frac {L}{9.8}$$
    $$\Rightarrow\, L\, =\, \displaystyle \frac {4\, \times\, 9.8}{4\, \times\, 3.14\, \times\, 3.14}\, m\, =\, 0.994\, m\, =\, 99.4\, cm$$
  • Question 4
    1 / -0
    The time period of a simple pendulum is 0.2 sec. What is its frequency of oscillation?
    Solution
    Frequency is inverse of time period.

    $$F= \dfrac { 1 }{ 0.2 } =5\ Hz$$
  • Question 5
    1 / -0
    You are designing a pendulum clock to have a period of $$1.0\ s$$. How long should the pendulum be ?
    Solution
    $$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$$ .. (1)

    Putting $$T = 1$$ in eqn. (1), 

    We get $$1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$$ $$\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$$ (approx.)
  • Question 6
    1 / -0
    What is to and fro motion of an object called?
  • Question 7
    1 / -0
    How does time period (T) of a seconds pendulum vary with length (L) ?
  • Question 8
    1 / -0
    A particle moving on x-axis has potential energy $$U=2-20x+5x^{2}$$ Joule along x-axis. The particle is released at $$r=-3$$. The maximum value of $$x$$ will be (x is in meter):
    Solution
    $$\displaystyle F=-\frac{dU}{dx}=20-10x$$
    $$F=0$$
    at $$x=0$$
    So, from $$x=-3$$ to $$x=2,$$ amplitude is $$5.$$  Hence other extreme position will be
    $$x=2+5=7$$
  • Question 9
    1 / -0
    A particle performing SHM takes time equal to T (time period of SHM) in consecutive appearances at a particular point. This point is 
    Solution
    In SHM, time period is defined as the time interval in which a particle returns to the same position after leaving that point. Now take one extreme position into consideration, then once particle leaves the position it comes again to the same position after a time period but in case of mid point it crosses two times. 
    Ans: extreme position
  • Question 10
    1 / -0
    Which of the following is not a necessary apparatus for the measurement of acceleration due to gravity $$(g)$$ by a simple pendulum?
    Solution
    In simple pendulum experiment, we require a stopwatch to measure the time taken for a certain number of oscillations, screw gauge to measure the diameter of the bob and scale to measure the length of the string. There is no need of Spherometer.
    Hence, Option C is correct.
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