A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand. (a) What is the amplitude of oscillation? (b) Find the frequency of oscillation?

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

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x’ is the extension in that case

F= kx’

F=mg

Mg=kx’

X’=mg/k

By dividing x by x’

x/x’= $\frac{2mg/k}{mg/k}=2$

so x=2x’

x’=4/2 =2cm

but the displacement of mass from the

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

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x’ is the extension in that case

F= kx’

F=mg

Mg=kx’

X’=mg/k

By dividing x by x’

x/x’= $\frac{2mg/k}{mg/k}=2$

so x=2x’

x’=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural length is equal to amplitude of the oscillation

A=x’=2cm

(b) Time period of the oscillating system depends on mass spring constant given by

T= $2\pi \sqrt[]{\frac{m}{k}}$

2mg/k=x

2mg/k=4cm =4 $\times {10}^{-2}m$

m/k =4 $\times {10}^{-2}$ /2g

k/m=g/2 $\times {10}^{-2}$

$v=\frac{1}{T}\sqrt[]{\frac{k}{m}}$

$v=\frac{1}{2\times 3.14}\sqrt[]{\frac{g}{2\times {10}^{-2}}}$

= $\frac{10}{628}\times 2.21$ =3.51Hz

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<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p><strong>(a)</strong> When the support of the hand is removed the body oscillates about mean position</p><p>Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745560009phpRF2j8M_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745560009phpRF2j8M.jpeg" alt="" width="223" height="131"></picture></div><div><p>Loss in PE of the block=mgx</p><p>Gain in elastic potential energy =1/2 kx²</p><p>By energy conservation we cam say that</p><p>Mgx=1/2kx²</p><p>Or x= 2mg/k</p><p>Now the mean position of oscillation will be when the block is balanced by spring</p></div></div><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745560067phpzgHbu1_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745560067phpzgHbu1.jpeg" alt="" width="126" height="167"></picture></div><div><p>If x’ is the extension in that case</p><p>F= kx’</p><p>F=mg</p><p>Mg=kx’</p><p>X’=mg/k</p><p>By dividing x by x’</p><p>x/x’=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>m</mi> <mi>g</mi> <mo>/</mo> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> <mi>g</mi> <mo>/</mo> <mi>k</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> </math> </span></p><p>so x=2x’</p><p>x’=4/2 =2cm</p><p>but the displacement of mass from the mean position when spring attains its natural length is equal to amplitude of the oscillation</p><p>A=x’=2cm</p><p><strong>(b) </strong>Time period of the oscillating system depends on mass spring constant given by</p><p>T=<span contenteditable="false"> <math> <mn>2</mn> <mi>π</mi> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p><p>2mg/k=x</p><p>2mg/k=4cm =4<span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> </span></p><p>m/k =4 <span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span>/2g</p><p>k/m=g/2<span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p><p><span contenteditable="false"> <math> <mi>v</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> </mfrac> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p><p><span contenteditable="false"> <math> <mi>v</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mo>×</mo> <mn>3.14</mn> </mrow> </mrow> </mfrac> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>g</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p><p>= <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>628</mn> </mrow> </mrow> </mfrac> <mo>×</mo> <mn>2.21</mn> </math> </span>=3.51Hz</p></div></div>

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