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
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.
Similar Questions
A student fixes two ends of a rope to two rigid supports A and B, placed at a distance of 15m. He then gives a transverse horizontal jerk to create a pulse in the rope which moves from A to B, B to A, repeatedly. The pulse dies after completing three, to and fro journeys from A to B and back. The pulse takes 2 min and 30s to complete its entire journey. The speed of the pulse moving on the rope is
A body moving in the positive x direction passes the origin at time t = 0. Between t = 0 and t = 1 second, thebody has a constant speed of 24 meters per second. At t = 1 second, the body is given a constant acceleration of6 meters per second squared in the negative x direction. The position x of the body at t = 11 seconds is(A) + 99m (B) + 36m (C) – 36 m (E) – 99 m
A body starting from rest accelerates uniformly by along a straight line at the rate of 10 m/s2 for 5s. It moves for 2s with uniform velocity of 50 m/s. Then it retards uniformly and comes to rest in 3s. The total distance travelled by the body is:A 200 m B 300 m C 400 m D 600 m
For the wave in the previous question, that was moving at 250250 m/s up,in which direction could the displacement of the individual particles of the string be occuring?A) left and rightB) up and downC) a combination of the twoD) impossible to tell
For the wave in the previous question, that was moving at 200200 m/s right,in which direction could the displacement of the individual particles of the string be occuring?A) left and rightB) up and downC) a combination of the twoD) impossible to tell
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.