A transverse wave along a string is given by y-2sin2-3t-x-4- where x and y are in cm and t is in second- The acceleration of a particle located at X - 4 cm at t - 1 sec is -

The correct option is D 36√(2)π^2cm/s^2 y=2 sin (6π t 2π x +π4) v_p=dydt=12π cos(6π t 2π x+π4) a_p=dv_pdt #160; #160; #160; ...

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Standard XII

Physics

Finding Average Velocity

A transverse ...

Question

A transverse wave along a string is given by $y = 2 sin (2 π (3 t - x) + \frac{π}{4})$ , where $x$ and $y$ are in $c m$ and $t$ is in $s e c o n d$ . The acceleration of a particle located at $X = 4 c m$ at $t = 1 s e c$ is :

36 \sqrt{2} π^{2} c m / s^{2}

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36 π^{2} c m / s^{2}

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- 36 \sqrt{2} π^{2} c m / s^{2}

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- 36 π^{2} c m / s^{2}

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Solution

The correct option is D $- 36 \sqrt{2} π^{2} c m / s^{2}$
$y = 2 sin (6 π t - 2 π x + \frac{π}{4})$
$v_{p} = \frac{d y}{d t} = 12 π cos (6 π t - 2 π x + \frac{π}{4})$
$a_{p} = \frac{d v_{p}}{d t}$
$= - 72 π^{2} sin (6 π t - 2 π x + \frac{π}{4})$

At $x = 4, t = 1$
$a_{p} = - 72 π^{2} sin (6 π - 8 π + \frac{π}{4})$
$= - 72 π^{2} \times \frac{1}{\sqrt{2}}$
$= - 36 \sqrt{2} π^{2} c m / s e c^{2}$

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A transverse wave along a string is given by y=2sin(2π(3t−x)+π4), where x and y are in cm and t is in second. The acceleration of a particle located at X=4 cm at t=1 sec is :

The correct option is D −36√2π2cm/s2y=2 sin(6πt−2πx+π4)vp=dydt=12π cos(6πt−2πx+π4)ap=dvpdt =−72π2 sin(6πt−2πx+π4)At x=4, t=1ap=−72π2 sin(6π−8π+π4) =−72π2×1√2 =−36√2π2 cm/sec2

A transverse wave along a string is given by $y = 2 sin (2 π (3 t - x) + \frac{π}{4})$ , where $x$ and $y$ are in $c m$ and $t$ is in $s e c o n d$ . The acceleration of a particle located at $X = 4 c m$ at $t = 1 s e c$ is :