NEET UG PHYSICS MCQ: OSCILLATIONS

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Two simple pendulums first of bob mass

{M_1}

and length

{L_1}

second of bob mass

{M_2}

and length

{L_2}

{M_1}

{M_2}

and

{L_1}

{2L_2}

. If the vibrational energy of both is same. Then which is correct

View answer and explanation

Correct option: B

Amplitude of B smaller than A

Option B, Amplitude of B smaller than A, is correct because for the same vibrational energy, the pendulum with a shorter length (and thus higher frequency) will have a smaller amplitude.

More Questions from OSCILLATIONS

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A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is

{f_1}

. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is

{f_2}

. Which of the following relations between the frequencies is correct The time period of a loaded spring on earth is 6s. On the surface of moon, the time period of the same loaded spring will be. Under the action of a force

F=-k x^3

, the motion of a particle is

(k=

a positive constant

)

S-I: A particle executes SHM with frequency

f

. The frequency with which total mechanical energy of particle oscillates is

2 f

.
S-II: Both kinetic as well as potential energy of particle in SHM oscillate with a frequency

2 f.

The phase difference (in radian) between two SHM whose equations are

y_1=10 \sin \left(10 \pi t+\frac{\pi}{3}\right)

and

y_2=12 \sin \left(8 \pi t+\frac{\pi}{4}\right)

t=0.5 \mathrm{~s}

is: A vertical spring block system is made to oscillate.
Statement-1: Its time period on earth is more than that on the moon.
Statement-2: Its extension on earth (in equilibrium) is more than that on the moon.