Find the time period of mass M when displaced from its equilibrium positon and then released for the system shown in Fig.

This is a short answer type question as classified in NCERT Exemplar

For calculation purpose, in this situation we will neglect gravity because it is constant throughout will not affect the net restoring force.

Let in the equilibrium position, the spring has extended by an amount x_o

Let displacement by spring is and string be x .

But string is extensible so only spring will contribute in extension  x+x=2x

So net extension is 2x+x_o

So force is F= 2T

T=kx_o

F=2kx_o

But when mass is lowered down further by x

F’=2T’ but spring length is 2x+x_o

F’=2k (2x+x_o)

Restoring force on the system

F_restoring=- (F’-F)

So using above equati

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This is a short answer type question as classified in NCERT Exemplar

For calculation purpose, in this situation we will neglect gravity because it is constant throughout will not affect the net restoring force.

Let in the equilibrium position, the spring has extended by an amount x_o

Let displacement by spring is and string be x .

But string is extensible so only spring will contribute in extension  x+x=2x

So net extension is 2x+x_o

So force is F= 2T

T=kx_o

F=2kx_o

But when mass is lowered down further by x

F’=2T’ but spring length is 2x+x_o

F’=2k (2x+x_o)

Restoring force on the system

F_restoring=- (F’-F)

So using above equations

F_restoring= [2k (2x+x₀)-2Kx_o]=-4kx

Ma=-4kx

So a is directly proportional to negative displacement so it follows SHM.

So time period is T= $\frac{2\pi }{w}$

w= $\sqrt[]{\frac{4k}{m}}$ so T= 2 $\pi \sqrt[]{\frac{M}{4K}}$

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<div><div><div><div><p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>For calculation purpose, in this situation we will neglect gravity because it is constant throughout will not affect the net restoring force.</p><p>Let in the equilibrium position, the spring has extended by an amount x_o</p><p>Let displacement by spring is and string be x .</p><p>But string is extensible so only spring will contribute in extension&nbsp; x+x=2x</p><p>So net extension is 2x+x_o</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745500793phpfQTpqj_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745500848phpNnHtRS_480x360.jpeg" alt="" width="136" height="138"></picture></div><div><p>So force is F= 2T</p><p>T=kx_o</p><p>F=2kx_o</p><p>But when mass is lowered down further by x</p><p>F’=2T’ but spring length is 2x+x_o</p><p>F’=2k (2x+x_o)</p><p>Restoring force on the system</p><p>F_restoring=- (F’-F)</p><p>So using above equations</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745500848phpNnHtRS_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745500848phpNnHtRS.jpeg" alt="" width="140" height="125"></picture></div><div><p>F_restoring= [2k (2x+x₀)-2Kx_o]=-4kx</p><p>Ma=-4kx</p><p>So a is directly proportional to negative displacement so it follows SHM.</p><p>So time period is T=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> </mfrac> </math> </span></p><p>w= <span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>4</mn> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span>so T= 2<span contenteditable="false"> <math> <mi>π</mi> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <mi>K</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p></div></div></div></div></div></div></div></div>

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