Question

The equation for a wave travelling in x-direction on a string is y = 3 sin [(3.14x - (314)t] where x is in cm and t is in second. (i) Find the maximum velocity of a particle of the string. (ii) Find the acceleration of a particle at x = 6.0 cm at time t = 0.11 s.

Answer 1

Explanation:

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Answer 2

Answer:

(1) maximum velocity is 9.42 m/sec

(2) acceleration = 0

Explanation:

The maximum velocity is v= $\dfrac{dy}{dt}$ and the acceleration is

a=$\frac{dv}{dt}$

The velocity of a particle at x at time t = $\dfrac{dy}{dt}$

we have equation y= 3 sin [3.14x - 314t]

Differentiating the given equation with respect to t, we get velocity

$\dfrac{dy}{dt}$ = 3(314) cos[3.14x - 314t]

$\dfrac{dy}{dt}$ = 942(cm/sec) cos[3.14x - 314t]

$\dfrac{dy}{dt}$ = 9.42 (m/sec) cos[3.14x - 314t]

and the maximum velocity is 9.42 m/sec

The acceleration of the particle at x=6.0 at time t=0.11 is  

a= $\dfrac{dv}{dt}$

a= $\dfrac{d(9.42 (m/sec) cos[3.14x - 314t])}{dt}$

a = - 9.42*314sin [3.14x - 314t]

a = -(2952 m/sec)sin [3.14x - 314t]

a = -(2952 m/sec)sin [3.14*6.0 - 314*0.11]

a = $-(2952 m/sec^{2})$sin[6π-11π]

a = 0