Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2° to the right with the vertical, the other pendulum makes an angle of 1° to the left of the vertical. What is the phase difference between the pendulums?

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

By considering the diagram

_{$\theta$ 1}= ${\theta }_{o}sin(wt+{\varnothing }_1)$

_{$\theta$ 2}= ${\theta }_o\mathrm{s}\mathrm{i}\mathrm{n}?\left(wt+{\varnothing }_2\right)$

As it is clear given that amplitude time period being equal but phases being different. Now for first pendulum at any time t

_{$\theta$ 1}= $\theta$ ₂

So sin $\frac{\pi }{2}$ = sin(wt+ ${\varnothing }_1$ )

wt+ ${\varnothing }_1=\frac{\pi }{2}$

where $\theta$ _o=2^o is the angular amplitude of first pendulum . for the second pendulum , the angular displacement is one degree , therefore $\theta$ ₂= $\frac{{\theta }_0}{2}$  and negative sign is taken to show for being left to mean position.

- $\frac{{\theta }_o}{2}={\theta }_{0}sin(wt+{\theta }_2)$

Sin(wt+ ${\varnothing }_2$ )=-1/2

So (wt+ $\theta$ ₂)=- $\frac{\pi }{6}or\frac{7\pi }{6}$

So by making their difference

(wt+ $\theta$ ₂)-( $wt+\theta$ ₁)=7 $\frac{\pi }{6}-\frac{\pi }{2}$ =4 $\frac{\pi }{6}$

( $\theta$ _{2- $\theta$ 1})= 120⁰

<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>By considering the diagram</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745558802phpdQKmol_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745558802phpdQKmol.jpeg" alt="" width="269" height="145"></picture></div><div><p>_{<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>1}=<span contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mo>(</mo> <mi>w</mi> <mi>t</mi> <mo>+</mo> <msub> <mrow> <mrow> <mi>∅</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>)</mo> </math> </span></p><p>_{<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>2}=<span contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">i</mi> <mrow> <mrow> <mi mathvariant="normal">n</mi> </mrow> <mo>?</mo> <mrow> <mfenced separators="|"> <mrow> <mrow> <mi>w</mi> <mi>t</mi> <mo>+</mo> <msub> <mrow> <mrow> <mi>∅</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </math> </span></p><p>As it is clear given that amplitude time period being equal but phases being different. Now for first pendulum at any time t</p><p>_{<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>1}= <span contenteditable="false"> <math> <mi>θ</mi> </math> </span>₂</p><p>So sin <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span>= sin(wt+ <span contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>∅</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> </span>)</p><p>wt+<span contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>∅</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>where <span contenteditable="false"> <math> <mi>θ</mi> </math> </span>_o=2^o is the angular amplitude of first pendulum . for the second pendulum , the angular displacement is one degree , therefore <span contenteditable="false"> <math> <mi>θ</mi> </math> </span>₂= <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span>&nbsp;and negative sign is taken to show for being left to mean position.</p><p>-<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mo>(</mo> <mi>w</mi> <mi>t</mi> <mo>+</mo> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>)</mo> </math> </span></p><p>Sin(wt+<span contenteditable="false"> <math> <msub> <mrow> <mrow> <mi>∅</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> </span> )=-1/2</p><p>So (wt+<span contenteditable="false"> <math> <mi>θ</mi> </math> </span> ₂)=-<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </mfrac> <mi>o</mi> <mi>r</mi> <mi> </mi> <mfrac> <mrow> <mrow> <mn>7</mn> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>So by making their difference</p><p>(wt+ <span contenteditable="false"> <math> <mi>θ</mi> </math> </span>₂)-(<span contenteditable="false"> <math> <mi>w</mi> <mi>t</mi> <mo>+</mo> <mi>θ</mi> </math> </span> ₁)=7 <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span>=4<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </mfrac> </math> </span></p><p>( <span contenteditable="false"> <math> <mi>θ</mi> </math> </span>_{2-<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>} ₁)= 120⁰</p></div></div>

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