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NEET UG PHYSICS MCQ: OSCILLATIONS

1 Easy

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Application +4 | -1

A second pendulum is moved to moon, where acceleration due to gravity is

1 / 6

times that of the earth, the length of the second pendulum on moon would be

View answer and explanation

Correct option: C

\frac{1}{6}

times

Option C (

\frac{1}{6}

times) is correct because the time period of a second pendulum is proportional to the square root of the length, and inversely proportional to the square root of the acceleration due to gravity. Since g is 1/6th on the moon, the length must also be 1/6th to maintain a 2-second period.

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)

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.