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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 upward passing through mean position. The pulse travels with speed of 3 m/s on the string & his hands passes 6 times in each second from the mean position. Then the point on the string at a distance 3 m will reach its upper extreme first time at time :- 1.25 sec. 1 sec  sec.none

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 upward passing through mean position. The pulse travels with speed of 3 m/s on the string & his hands passes 6 times in each second from the mean position. Then the point on the string at a distance 3 m will reach its upper extreme first time at time :- 1.25 sec. 1 sec  sec.none

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Solution

The question is asking for the time it takes for a point on the string, 3 meters away, to reach its upper extreme for the first time.

Given that the speed of the pulse is 3 m/s, it would take 1 second for the pulse to reach a point 3 meters away (since speed = distance/time, so time = distance/speed = 3m/3m/s = 1s).

However, the man's hand passes through the mean position 6 times per second, which means the period of the motion (time for one complete cycle) is 1/6 seconds.

Since the hand is moving upwards from the mean position at t=0, it will reach the upper extreme at t = 1/4 of the period (since it's a symmetrical pulse and the hand is at the mean position at t=0, it will take a quarter of the period to reach the upper extreme).

So, the time it takes for the hand to reach the upper extreme is 1/4 * 1/6 = 1/24 seconds.

Adding these two times together, the point on the string 3 meters away will reach its upper extreme for the first time at t = 1s (time for pulse to reach the point) + 1/24s (time for hand to reach upper extreme) = 1.04167 seconds.

So, the answer is none of the given options.

This problem has been solved

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