Question

BEGINNER'S BOX-2
A travelling wave in stretched string is given by the equation : y = 40 cos(3x - 5t) cm where tis in sec. Determine
the maximum speed of particles of medium.​

Answer 1

Answer:

First of all, you need to see the proof of velocity of wave.

let v = velocity, f = frequency , lambda be wavelength.

Now, v = f* lambda

v = f* lambda

v = f* lambda => v = (2πf)/(2π/lambda)

inserting 2π in numerator and denominator.

=> v = (w/k),

where w = 2πf and k = 2π/k

in other words, w is the coefficient of "t" and k is the coefficient of "x" in the given equation of the question.

w in the given equation is 5 and k is 3

Taking this equation,

V = 5/3 m/s.

This velocity represents the velocity of wave.

For velocity of particles, differentiate the given equation.

y = 40 cos (3x-5t)

=> dy/dt = 40*5 [- sin(3x-5t)]

therefore velocity of medium particles is given by the above equation.

v = 200[-sin(3x-5t)]

for velocity to be maximum, value of

value of sin(3x-5t) should be 1

Therefore V max = 200 m/sec.

This is velocity of medium particles.