One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

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

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $\rho =$ density of the liquid

PE= dmgx=A $\rho dxgx$

So total PE of the column

= ${\int }_0^{h_1}Adxg\rho x$ = $Ag\rho {\int }_o^{h_1}xdx=A\rho g\frac{h_1^2}{2}$

But h₁=lsin45

PE=A $\rho$ gl²sin²45/2

Similarly PE of right column = A $\rho$ gl²sin²45/2

Total PE = A $\rho$ gl²sin²45/2+ A $\rho$ gl²sin²45/2

= A $\rho$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A\rho g}{2}[{\left(l-y\right)}^2+{\left(l+y\right)}^2-l^2$ ]

= $\frac{A\rho g}{2}[2{(l}^2+y^2)]$ =A $\rho g{(l}^2+y^2)$

Change in KE = ½ mv²

m=A $\rho 2l$

change in KE= 1/2A $\rho 2l{v}^2$ =A $\rho l{v}^2$

s

...more

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

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $\rho =$ density of the liquid

PE= dmgx=A $\rho dxgx$

So total PE of the column

= ${\int }_0^{h_1}Adxg\rho x$ = $Ag\rho {\int }_o^{h_1}xdx=A\rho g\frac{h_1^2}{2}$

But h₁=lsin45

PE=A $\rho$ gl²sin²45/2

Similarly PE of right column = A $\rho$ gl²sin²45/2

Total PE = A $\rho$ gl²sin²45/2+ A $\rho$ gl²sin²45/2

= A $\rho$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A\rho g}{2}[{\left(l-y\right)}^2+{\left(l+y\right)}^2-l^2$ ]

= $\frac{A\rho g}{2}[2{(l}^2+y^2)]$ =A $\rho g{(l}^2+y^2)$

Change in KE = ½ mv²

m=A $\rho 2l$

change in KE= 1/2A $\rho 2l{v}^2$ =A $\rho l{v}^2$

so from above eqn

change in PE +change in KE = A $\rho g{(l}^2+y^2)$ +A $\rho l{v}^2$

system being conservative

so change in total energy =0

A $\rho g{(l}^2+y^2)+A\rho l{v}^2$ =0

Differentiating both with respect to time t

A $\rho g\left(2yv\right)+A\rho l\left(2v\right)a=0$ =0

Dy/dt =v and dv/dt=a

A

(gy+la)2A $\rho v$ =0

2A $\rho v=constant$ and 2A $\rho v\ne 0$

La+gy=0

A+(gy/l)=0

d²y/dt²+gy/l=0

d²y/dt²+w²y=0

w= $\sqrt[]{\frac{g}{l}}$

T=2 $\pi \sqrt[]{\frac{l}{g}}$

less

<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.</p><div><div><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745560402phptpcKEl_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745560402phptpcKEl.jpeg" alt="" width="204" height="138"></picture></div><div><p>If<span contenteditable="false"> <math> <mi>ρ</mi> <mo>=</mo> </math> </span> density of the liquid</p><p>PE= dmgx=A<span contenteditable="false"> <math> <mi>ρ</mi> <mi>d</mi> <mi>x</mi> <mi>g</mi> <mi>x</mi> </math> </span></p><p>So total PE of the column</p><p>= <span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <msub> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </msubsup> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mi>d</mi> <mi>x</mi> <mi>g</mi> <mi>ρ</mi> <mi>x</mi> </math> </span>=<span contenteditable="false"> <math> <mi>A</mi> <mi>g</mi> <mi>ρ</mi> <msubsup> <mo>∫</mo> <mrow> <mi>o</mi> </mrow> <mrow> <msub> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </msubsup> <mrow> <mrow> <mi>x</mi> <mi>d</mi> <mi>x</mi> </mrow> </mrow> <mo>=</mo> <mi>A</mi> <mi>ρ</mi> <mi>g</mi> <mfrac> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>But h₁=lsin45</p><p>PE=A <span contenteditable="false"> <math> <mi>ρ</mi> </math> </span>gl²sin²45/2</p><p>Similarly PE of right column = A <span contenteditable="false"> <math> <mi>ρ</mi> </math> </span>gl²sin²45/2</p><p>Total PE = A <span contenteditable="false"> <math> <mi>ρ</mi> </math> </span>gl²sin²45/2+ A <span contenteditable="false"> <math> <mi>ρ</mi> </math> </span>gl²sin²45/2</p><p>= A <span contenteditable="false"> <math> <mi>ρ</mi> </math> </span>gl²/2</p><p>If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y</p><p>But liquid present in the right arm =l+y</p><p>Total PE = PE_final-PE_initial</p><p>Change in PE =<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>A</mi> <mi>ρ</mi> <mi>g</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>[</mo> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mi>l</mi> <mo>-</mo> <mi>y</mi> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mi>l</mi> <mo>+</mo> <mi>y</mi> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span> ]</p><p>= <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>A</mi> <mi>ρ</mi> <mi>g</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>[</mo> <mn>2</mn> <msup> <mrow> <mrow> <mo>(</mo> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>)</mo> <mo>]</mo> </math> </span>=A<span contenteditable="false"> <math> <mi>ρ</mi> <mi>g</mi> <msup> <mrow> <mrow> <mo>(</mo> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>)</mo> </math> </span></p><p>Change in KE = ½ mv²</p><p>m=A<span contenteditable="false"> <math> <mi>ρ</mi> <mn>2</mn> <mi>l</mi> </math> </span></p><p>change in KE= 1/2A<span contenteditable="false"> <math> <mi>ρ</mi> <mn>2</mn> <mi>l</mi> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span> =A<span contenteditable="false"> <math> <mi>ρ</mi> <mi>l</mi> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p><p>so from above eqn</p><p>change in PE +change in KE = A<span contenteditable="false"> <math> <mi>ρ</mi> <mi>g</mi> <msup> <mrow> <mrow> <mo>(</mo> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>)</mo> </math> </span> +A<span contenteditable="false"> <math> <mi>ρ</mi> <mi>l</mi> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p><p>system being conservative</p><p>so change in total energy =0</p><p>A <span contenteditable="false"> <math> <mi>ρ</mi> <mi>g</mi> <msup> <mrow> <mrow> <mo>(</mo> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>)</mo> <mo>+</mo> <mi>A</mi> <mi>ρ</mi> <mi>l</mi> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span>=0</p><p>Differentiating both with respect to time t</p><p>A <span contenteditable="false"> <math> <mi>ρ</mi> <mi>g</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <mi>y</mi> <mi>v</mi> </mrow> </mrow> </mfenced> <mo>+</mo> <mi>A</mi> <mi>ρ</mi> <mi>l</mi> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <mi>v</mi> </mrow> </mrow> </mfenced> <mi>a</mi> <mo>=</mo> <mn>0</mn> </math> </span>=0</p><p>Dy/dt =v and dv/dt=a</p><p>A</p><p>(gy+la)2A <span contenteditable="false"> <math> <mi>ρ</mi> <mi>v</mi> </math> </span>=0</p><p>2A <span contenteditable="false"> <math> <mi>ρ</mi> <mi>v</mi> <mo>=</mo> <mi>c</mi> <mi>o</mi> <mi>n</mi> <mi>s</mi> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>t</mi> <mi> </mi> </math> </span>and 2A<span contenteditable="false"> <math> <mi>ρ</mi> <mi>v</mi> <mo>≠</mo> <mn>0</mn> </math> </span></p><p>La+gy=0</p><p>A+(gy/l)=0</p><p>d²y/dt²+gy/l=0</p><p>d²y/dt²+w²y=0</p><p>w=<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>g</mi> </mrow> </mrow> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p><p>T=2 <strong><span contenteditable="false"> <math> <mi>π</mi> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mi>g</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></strong></p></div></div></div></div>

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