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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 |
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