A transverse wave along a string is given by y=2sin (2 π (3 t - x) + (π /4)), where, x and y are in cm and t in second. Find the acceleration of a particle located at x=4 cm at t=1 s .

Question

A transverse wave along a string is given by y=2sin (2 π (3 t - x) + (π /4)), where, x and y are in cm and t in second. Find the acceleration of a particle located at x=4 cm at t=1 s . (A) 36√2 π 2 cms- 2 (B) 36π 2 cms- 2 (C) -36√2 π 2 cm s- 2 (D) -36π 2 cms- 2

Tardigrade Admin · Accepted Answer

We know that v=(d y/d t) =(d/d t) 2sin [2 π (3 t - x) + (π /4)] a=(d v/d t)=(d/d t)12π cos (2 π (3 t - x) + (π /4)) a=-72 π2 sin (2 π(3 t-x)+(π/4)) at t=1 and x=4 cm a=-36 √2π 2 cm s- 2

Q. A transverse wave along a string is given by $y = 2 s in (2 π (3 t - x) + \frac{π}{4}),$ where, $x$ and $y$ are in $c m$ and $t$ in second. Find the acceleration of a particle located at $x = 4 c m$ at $t = 1 s$ .

$362 π^{2} c m s^{- 2}$

$36 π^{2} c m s^{- 2}$

$- 362 π^{2} c m s^{- 2}$

$- 36 π^{2} c m s^{- 2}$

We know that $v = \frac{d y}{d t}$
$= \frac{d}{d t} 2 s in [2 π (3 t - x) + \frac{π}{4}]$
$a = \frac{d v}{d t} = \frac{d}{d t} 12 π cos (2 π (3 t - x) + \frac{π}{4})$
$a = - 72 π^{2} sin (2 π (3 t - x) + \frac{π}{4})$
at $t = 1$ and $x = 4 c m$
$a = - 362 π^{2} c m s^{- 2}$

Q. A transverse wave along a string is given by y=2sin(2π(3t−x)+4π​), where, x and y are in cm and t in second. Find the acceleration of a particle located at x=4cm at t=1s .

362​π2cms−2

36π2cms−2

−362​π2cms−2

−36π2cms−2