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

Given :

A transverse wave along a string = $\sf y = 2 sin \bigg( 2\pi (3t - x) + \dfrac{\pi}{4}\bigg)$

x = 4cm

t = 1 cm

To find :

acceleration of the particle.

Solution :

The acceleration of a particle at displacement is given by a = -ω²y

According to the question,

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

$\sf \omega = 6\pi t$

By substituting all the given values in the formula,

$\dashrightarrow \sf a = - \omega ^{2} y$

$\dashrightarrow \sf a = - (6\pi t) ^{2} \times 2 \: sin \bigg(6\pi t - 2\pi x + \dfrac{\pi}{4} \bigg)$

$\dashrightarrow \sf a = - 72\pi ^{2} (1) \times 2 \: sin \bigg(6\pi (1) - 2\pi (4) + \dfrac{\pi}{4} \bigg)$

$\dashrightarrow \sf a = - 72\pi ^{2} \times 2 \: sin \bigg(6\pi - 8\pi + \dfrac{\pi}{4} \bigg)$

$\dashrightarrow \sf a = - 72\pi ^{2} \times 2 \: sin \bigg( - 2\pi+ \dfrac{\pi}{4} \bigg)$

$\dashrightarrow \sf a = - 72\pi ^{2} \times 2 \: sin \bigg( \dfrac{ - 8\pi+\pi}{4} \bigg)$

$\dashrightarrow \sf a = - 72\pi ^{2} \times \: sin ( - 315)$

$\dashrightarrow \sf a = - 72\pi ^{2} \times \: ( - sin 45)$

$\dashrightarrow \sf a = 72\pi ^{2} \dfrac{1}{ \sqrt{2} }$

$\dashrightarrow \underline{\boxed{\sf a = 36 \sqrt{2} {\pi}^{ 2 }cms^{ - 2} }}$

Thus, acceleration of the particle is 36√2 π² cm/s²