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NEET UG PHYSICS MCQ: OSCILLATIONS

1 Easy

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Application +4 | -1

The length of a seconds pendulum at a height

h=2 R

from earth surface will be
(Given

R=

Radius of earth and acceleration due to gravity at the surface of earth,

g=\pi^2 \mathrm{~ms}^{-2}

)

View answer and explanation

Correct option: D

\frac{1}{9} \mathrm{~m}

Option D (

\frac{1}{9} m

) is correct because the acceleration due to gravity decreases with height, affecting the length of the seconds pendulum.

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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.