Question

A man generates a symmetrical pulse in a string by moving his hand up and down. At t=0, the point in his hand moves downwards from mean position. The pulse travels with speed 3 m/s on the string and his hand passes 6 times in each second from the mean position. Then the point of the string at a distance 3 m will reach its upper extreme first time at t=

Answer 1

Firstly, we have to draw the pulse wave (see attached diagram).

If hand passes 6 times from the mean position in one second, then we know that string creates 3 wave lengths (λ) or 3 cycles after 1 second.

That means frequency (f) of the wave is 3 Hz.

Now we can use below equation to calculate the value of wave length.

V = fλ (V = velocity of the wave)

λ = V/f

  = (3m/s)/3 = 1 m

If λ = 1m, the point which have 3m distance is located at no.(6) (in the diagram).

to reach its upper extreme ----> have to travel 3λ/4 distance

time to travel 3λ = 1 second

time to travel λ = 1/3 seconds

time to travel 3λ/4 = (1/3) * (3/4) seconds

                               = 1/4 seconds = 0.25 seconds

Answer :0.25 seconds

Answer 2

Answer:

ans is t=3/3+3T/4 t=1/v =2/6=1/3

t=1+1/4 = 1.25

I hope it helps you

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