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
B
34
C
43
D
169

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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