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)

    \frac{1}{4} Hz

  • Option 2)

    \frac{1}{2\sqrt{2}}Hz

  • Option 3)

    \frac{1}{2}Hz

  • Option 4)

    2 Hz

 

Answers (1)

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)

\frac{1}{4} Hz

This is an incorrect option.

Option 2)

\frac{1}{2\sqrt{2}}Hz

This is an incorrect option.

Option 3)

\frac{1}{2}Hz

This is the correct option.

Option 4)

2 Hz

This is an incorrect option.

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