The vibration of a string fixed at both ends | vedclass

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(D) Given the wave equation: $Y = 2 \sin(\pi x) \sin(100\pi t)$.For option $(A)$: The displacement of a particle at position $x$ is $Y(t) = [2 \sin(\p

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Both $(A)$ and $(B)$.

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(D) Given the wave equation: $Y = 2 \sin(\pi x) \sin(100\pi t)$.For option $(A)$: The displacement of a particle at position $x$ is $Y(t) = [2 \sin(\pi x)] \sin(100\pi t)$. The amplitude is $A = |2 \sin(\pi x)|$. At $x = 1/6 \, cm$,$A = 2 \sin(\pi/6) = 2 	imes 0.5 = 1 \, mm$. Thus,$(A)$ is correct.For option $(B)$: The velocity of the particle is $v = \frac{\partial Y}{\partial t} = 2 \sin(\pi x) 	imes 100\pi \cos(100\pi t) = 200\pi \sin(\pi x) \cos(100\pi t)$.At $x = 1/6 \, cm$ and $t = 1/600 \, sec$:$v = 200\pi \sin(\pi/6) \cos(100\pi/600) = 200\pi 	imes (1/2) 	imes \cos(\pi/6) = 100\pi 	imes (\sqrt{3}/2) = 50\pi\sqrt{3} \, mm/s$.Since $\pi \approx 3.14$,$50 	imes 3.14 	imes \sqrt{3} = 157\sqrt{3} \, mm/s$. Thus,$(B)$ is correct.For option $(C)$: The wave number $k = \pi$. Since $k = \frac{n\pi}{L}$,we have $\pi = \frac{n\pi}{10}$,which gives $n = 10$. Thus,$(C)$ is incorrect.

The vibration of a string fixed at both ends is described by $Y = 2 \sin(\pi x) \sin(100\pi t)$,where $Y$ is in $mm$,$x$ is in $cm$,and $t$ is in $sec$. Then:

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