Question

A transverse wave along na string is given by

$\sf y = 2 sin \bigg( 2\pi (3t - x) + \dfrac{\pi}{4}\bigg)$

where X and y are in cm and t is in second. the acceleration of a particle located at x = 4cm at t = 1s​

Answer 1

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The type of wave that occurs in a string is called a transverse wave. In a transverse wave, the wave direction is perpendicular the the direction that the string oscillates in. ... The speed of a wave is proportional to the wavelength and indirectly proportional to the period of the wave: v=λT v = λ T .

Step-by-step explanation:

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Answer 2

Answer:

• The Acceleration of particle is 36√2π² cm/s²

Explanation:

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Given equation,

$\longrightarrow\sf y=2\sin\Bigg(2\pi(3t-x)+\dfrac{\pi}{4}\Bigg)\\\\\\\\\longrightarrow\sf y=2\sin\Bigg(6\pi t-2\pi x+\dfrac{\pi}{4}\Bigg)$

Now, we know that for velocity v = dy/dt .

$\longrightarrow\sf v = \dfrac{d}{dt} \Bigg[2\sin\Bigg(6\pi t-2\pi x+\dfrac{\pi}{4}\Bigg)\Bigg]$

Applying chain rule,

$\longrightarrow\sf v = \bigg(2\times 6\pi\bigg)\times \cos\Bigg(6\pi t-2\pi x+\dfrac{\pi}{4}\Bigg)\\\\\\\\\longrightarrow\underline{\sf v =12\pi \; \cos\Bigg(6\pi t-2\pi x+\dfrac{\pi}{4}\Bigg)}$

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Similarly to get acceleration, a = dv/dt .

$\longrightarrow\sf a = \dfrac{d}{dt} \Bigg[12\pi\;\cos\Bigg(6\pi t-2\pi x+\dfrac{\pi}{4}\Bigg)\Bigg]$

Similarly applying chain rule, we get

$\longrightarrow\sf a = \bigg(12\pi\times 6\pi\bigg)\times -\sin\Bigg(6\pi t-2\pi x+\dfrac{\pi}{4}\Bigg)\\\\\\\\\longrightarrow\underline{\sf a =-72\pi^{2} \; \sin\Bigg(6\pi t-2\pi x+\dfrac{\pi}{4}\Bigg)}$

Acceleration at x= 4 and t = 1 s will be,

$\longrightarrow\sf a =-72\pi^{2} \; \sin\Bigg(6\pi \times 1-2\pi \times 4+\dfrac{\pi}{4}\Bigg)\\\\\\\\\longrightarrow\sf a = -72\pi^{2} \; \sin\Bigg(6\pi -8\pi +\dfrac{\pi}{4}\Bigg)\\\\\\\\\longrightarrow\sf a = -72\pi^{2} \; \sin\Bigg(-2\pi +\dfrac{\pi}{4}\Bigg)$

From the identity sin(2π + θ) = sin θ

$\longrightarrow\sf a = -72\pi^{2} \; \sin\Bigg(\dfrac{\pi}{4}\Bigg)\\\\\\\\\longrightarrow\sf a = -72 \pi^{2}\times \dfrac{1}{\sqrt{2}}\\\\\\\\\longrightarrow\underline{\boxed{\sf a = -36\sqrt{2} \pi^{2}\;cm/s^{2}}}$

∴ The Acceleration of particle is 36√2π² cm/s².

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