Physics, asked by Aquamarine2568, 11 months ago

A weightless spring which has a force constant 'k' oscillates with frequency 'n' when a mass m is suspended from it. The spring is cut into two equal halves and a mass 2m is suspended from it. The frequency of oscillation will now become- a skipped n b 2n c n/ d n(2)1/2

Answers

Answered by KailashHarjo
5

Given,

a weightless spring which has a force constant 'k' oscillates with frequency 'n' when a mass m is suspended from it.

The period of oscillation of mass m of a body suspended by a spring is ,

T=2π(√m/k)

Also given that,

The spring is cut into two equal halves and a mass 2m is suspended from it.

So,

k2= 2k

m2=2m

t=1/n

→n=(1/2π) * √k/m

n2= 1/ 2π *√2k/2m

    =  (1/2π) * √k/m

    =n

Therefore,the new frequency is same as previous frequency.

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