1.

The frequency of oscillations of a mass m suspended by a spring is f1. If the length of the spring is cut to one-half, the same mass oscillates with frequency f2. Determine the value \(\frac{f_2}{f_1}\).

Answer»

Let the full spring be the combination of two springs in series, each of force constant k. then in case (I), the effective spring constant (k1) is given by

k1 = \(\frac{k\times k}{k\times k}=\frac{k}{2}\) 

Frequency of oscillation,

f1 = \(\frac{1}{2\pi}\sqrt{\frac{k_1}{m}}\)

= \(\frac{1}{2\pi}\sqrt{\frac {k}{2m}}\)

In case (ii); when the spring is cut to one-half, the effective spring constant k2 = k. Frequency of oscillation,

f2 = \(\frac{1}{2\pi}\sqrt{\frac{k_2}{m}}\)

= \(\frac{1}{2\pi}\sqrt{\frac{k}{m}}\) 

\(\frac{f_2}{f_1}=\sqrt 2\)

The frequency of oscillations of mass m and force constant k is, f1​ = 1/2π under root k/​m​​
when the spring is cut to one-half of its length, its force constant is doubles (2k).
Then frequency of oscillation of mass m will be 

f2​=1/2π under root ​2k/m​​ =​ f2/f1 ​​= under root2​`



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