NEET UG PHYSICS MCQ: 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

View answer and explanation

Correct option: D

{f_2} = \sqrt 2 {f_1}

Option D (

{f_2} = \sqrt 2 {f_1}

) is correct because cutting the spring in half doubles the spring constant, which increases the frequency by a factor of

\sqrt{2}

More Questions from OSCILLATIONS

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F=-k x^3

(k=

)

f

2 f

2 f.

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}

U\left( x \right) = k{\left[ x \right]^3}