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Question

A body of mass $$m$$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $$m$$ is slightly pulled down and released, it oscillates with a time period of $$3s$$. when the mass m is increased by $$1$$ kg the time period of oscillations becomes $$5s$$. Then value of $$m$$ in kg is


A
916
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B
34
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C
43
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D
169
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Solution

The correct option is A $$\dfrac{9}{16}$$
For SHM of a hanging mass by a spring,
$$F=-k_sx$$
where $$k_s=m\omega^2$$
$$\implies \omega=\sqrt{\dfrac{k_s}{m}}$$
Hence time period of SHM=$$T=\dfrac{2\pi}{\omega}$$
$$=2\pi\sqrt{\dfrac{m}{k_s}}\propto \sqrt{m}$$
Hence $$\dfrac{5}{3}=\sqrt{\dfrac{m+1}{m}}$$
$$\implies m=\dfrac{9}{16}kg$$

Physics

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