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Oscillations MCQs for NEET — Physics Questions with Answers

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The potential energy of a particle with displacement X depends as U(X). The motion is simple harmonic, when (K is a positive constant)

$U = \frac{K X^{2}}{2}$

$U = K X^{2}$

$U = K$

$U = K X$

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Explanation

(a) $F = - k x \Rightarrow d W = F d x = - k x d x$

$s o \int_{o}^{w} d W = \int_{0}^{x} - k x d x \Rightarrow W = - U = \frac{1}{2} k x^{2}$

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The angular velocity and the amplitude of a simple pendulum is $ω$ and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is

$X^{2} ω^{2} (a^{2} - X^{2} ω^{2})$

$X^{2} / (a^{2} - x^{2})$

$(a^{2} - X^{2} ω^{2}) / X^{2} ω^{2}$

$(a^{2} - x^{2}) / X^{2}$

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Explanation

(d) Kinetic energy $T = \frac{1}{2} m ω^{2} (a^{2} - x^{2})$

and potential energy $V = \frac{1}{2} m ω^{2} x^{2} s o \frac{T}{V} = \frac{a^{2} - x^{2}}{x^{2}}$

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A particle is executing simple harmonic motion with frequency f. The frequency at which its kinetic energy changes into potential energy, is:

f/2

2 f

4 f

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Explanation

(c) In SHM, KE changes into PE two times during one oscillation. So if the frequency of oscillations is f, then the frequency at which KE changes into PE is 2f.

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There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force , $F = - K x$ where x is the displacement. The total energy of body depends upon -

K, x

K, a

K, a, x

K, a, v

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Explanation

(b) Total energy $U = \frac{1}{2} K a^{2}$

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A simple pendulum is set up in a trolley which moves to the right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle $θ$ with the vertical

$\tan^{- 1} (\frac{a}{g})$ in the forward direction

$\tan^{- 1} (\frac{a}{g})$ in the upward direction

$\tan^{- 1} (\frac{a}{g})$ in the backward direction

$\tan^{- 1} (\frac{g}{a})$ in the forward directions

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The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)

$\frac{1}{8} E$

$\frac{1}{4} E$

$\frac{1}{2} E$

$\frac{2}{3} E$

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Explanation

(b)

$\frac{U}{E} = \frac{\frac{1}{2} m ω^{2} y^{2}}{\frac{1}{2} m ω^{2} a^{2}} = \frac{y^{2}}{a^{2}} \Rightarrow \frac{U}{E} = \frac{{(\frac{1}{2} a)}^{2}}{a^{2}} = \frac{1}{4} \Rightarrow U = \frac{E}{4} J$

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A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?

P.E. is maximum when x = 0

K.E. is maximum when x = 0

T.E. is zero when x = 0

K.E. is maximum when x is maximum

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Explanation

(b) In S.H.M., at mean position i.e. at x = 0 kinetic energy will be maximum and P.E. will be minimum. Total energy is always constant.

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The total energy of a particle, executing simple harmonic motion is

$\propto x$

$\propto x^{2}$

Independent of x

$\propto x^{1 / 2}$

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Explanation

(c)Total energy = $\frac{1}{2} m ω^{2} a^{2}$ = constant

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A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by $T = 2 π \sqrt{\frac{l}{g'}}$ , where $g^{'}$ is equal to

g-a

g+a

$\sqrt{g^{2} + a^{2}}$

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If the length of second's pendulum is decreased by 2%, how many seconds it will lose per day?

3927 sec

3727 sec

3427 sec

864 sec

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Explanation

$T \propto \sqrt{l} \Rightarrow \frac{∆ T}{T} = \frac{1}{2} \frac{∆ l}{l} = \frac{0.02}{2} = 0.01 Also, time measured by pendulum t = nT where n is number of oscillations \frac{∆ t}{t} = \frac{∆ T}{T} = 0.01$

Loss of time per day = $0.01 \times 24 \times 60 \times 60 = 864 \sec$

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