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A 4\mathrm{~kg} block attached to a spring vibrates with a frequency of 4 \mathrm{HZ} on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 12 \mathrm{~kg} lelock placed on the same table. So, the frequency (in Hz) of vibration of the 12 \mathrm{~kg} block

Option: 1

1.45


Option: 2

1.67


Option: 3

1.24


Option: 4

1.64


Answers (1)

best_answer

 n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=4 H z \\

k=4 \pi^2 \times m \times 4 \\

k=4 \pi^2 \times 4 \times 4 \\

k=64 \pi^2 \mathrm{~N} / \mathrm{m}

When two springs are attached in Parallel to an 12kg block, then

k_{e q}=k+k=2 k \\

n^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{k e q}{m^{\prime}}} \\

n^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{2 \times 64 \pi^2}{12}}=\frac{8 \pi}{2 \pi} \times \frac{1}{\sqrt{6}}=\frac{4}{\sqrt{6}}

n^{\prime}=\frac{4}{2.45}=\frac{400}{245}=1.64

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