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Question: A mass m is suspended from a string of length l and force constant k. The frequency of vibration of ...

A mass m is suspended from a string of length l and force constant k. The frequency of vibration of the mass is f1. The spring is cut in to two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is f2. Which of the following reaction between the frequencies is correct.

A

f1=2f2f _ { 1 } = \sqrt { 2 } f _ { 2 }

B

f1=f2f _ { 1 } = f _ { 2 }

C

f1=2f2f _ { 1 } = 2 f _ { 2 }

D

f2=2f1f _ { 2 } = \sqrt { 2 } f _ { 1 }

Answer

f2=2f1f _ { 2 } = \sqrt { 2 } f _ { 1 }

Explanation

Solution

f∝kf \propto \sqrt { k }

If the spring is divided in to equal parts then force constant of each part will becomes double

f2f1=k2k1=2\frac { f _ { 2 } } { f _ { 1 } } = \sqrt { \frac { k _ { 2 } } { k _ { 1 } } } = \sqrt { 2 } ⇒ f2=2f1f _ { 2 } = \sqrt { 2 } f _ { 1 }