A weightless spring which has a force constant 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

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Solution

The correct option is A
n

n = $\frac{1}{2 π} \sqrt{\frac{k}{m}} \Rightarrow \frac{n}{n^{'}} = \sqrt{\frac{k}{m} \times \frac{m^{'}}{K^{'}}} = \sqrt{\frac{k}{m} \times \frac{2 m}{2 K}} = 1 \Rightarrow n^{'} = n$

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A weightless spring which has a force constant 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

The correct option is A n n = 12π√km⇒nn′=√km×m′K′=√km×2m2K=1⇒n′=n

The correct option is A
n

n = $\frac{1}{2 π} \sqrt{\frac{k}{m}} \Rightarrow \frac{n}{n^{'}} = \sqrt{\frac{k}{m} \times \frac{m^{'}}{K^{'}}} = \sqrt{\frac{k}{m} \times \frac{2 m}{2 K}} = 1 \Rightarrow n^{'} = n$