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  • A system is shown in the figure. The time period for small oscillations of the two blocks will be.                         &nbs

A system is shown in the figure. The time period for small oscillations of the two blocks will be.

                                            

Option: 1

2\pi \sqrt{\frac{3\: m}{k}}         


Option: 2

2\pi \sqrt{\frac{3\: m}{2\: k}}


Option: 3

2\pi \sqrt{\frac{3\: m}{4\: k}}


Option: 4

2\pi \sqrt{\frac{3\: m}{8\: k}}


Answers (1)

best_answer

 

Time period of oscillation for spring mass system -

T= 2\pi \sqrt{\frac{m}{K}}

- wherein

m = mass of block

K = spring constant

 

 

Series combination of spring -

- wherein

\frac{1}{K_{eq}}= \frac{1}{K_{1}}+ \frac{1}{K_{2}}

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

 

 

Here, both the spring are in series

\therefore\; \; K_{eq}\; \; =\frac{K\left ( 2K \right )}{K+2K}=\frac{2K}{3}

Time period

T=2\pi \sqrt{\frac{\mu }{K_{eq}}} Where \mu =\frac{m_{1}m_{2}}{m_{1}+m_{2}}

Here \mu =\frac{m}{2} \therefore \; T=2\pi \sqrt{\frac{m}{2}\times \frac{3}{2K}}=2\pi \sqrt{\frac{3m}{4K}}

Method II

 \therefore mx_{1}=mx_{2}\Rightarrow\; x_{1}=x_{2}

force equation for first block;

\frac{2k}{3}\left ( x_{1}+x_{2} \right )=-m\frac{d^{2}x_{1}}{dt^{2}}

             

Put   x_{1}=x_{2}\Rightarrow \frac{d^{2}x_{1}}{dt^{2}}+\frac{4k}{3m}\times x_{1}=0\Rightarrow \omega ^{2}=\frac{4k}{3m}

\therefore \; T=2\pi \sqrt{\frac{3m}{4k}}

 

 

 

Posted by

Sanket Gandhi

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