# A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table.  Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is : Option 1) Option 2) Option 3) Option 4)

As we learnt in

Parallel combination of spring -

- wherein

$K_{eq}=K_{1}+K_{2}$

$K_{1}and\ K_{2}$ are spring constants of spring 1 & 2 respectively.

We know that $T=2\pi \sqrt{\frac{m}{k}}\ \; \Rightarrow\ \; \upsilon=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

$t=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\ \; \Rightarrow\ \; 4\pi^{2}=\frac{k}{m}\ \; \Rightarrow\ \; K=4\pi^{2}\ \; [\because\ \; m=1kg]$

If spring attached in Parallel. $K_{r}=2K\ \; m=8kg$

$\upsilon=\frac{1}{2\pi}\sqrt{\frac{2\times 4\pi^{2}}{8}}\ \; \Rightarrow\ \; \upsilon=\frac{1}{2}Hz$

Correct option is 3.

Option 1)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is the correct option.

Option 4)

This is an incorrect option.

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