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A strings is stretched and a pulse is created along it. The stopwatch is started from its position `A`, when the pulse is in position `X` and is stopped in its position `B`, when the pulse has travelled back to its position `X`. Find the velocity of propagation of the pulse along the string. . |
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Answer» Initial reading of stopwatch (position A) `= 55 s` Final reading of stopwatch (position B) `= 1 min 45 s = 105 s` Time interval `= 105 s - 55 s = 50 s` Distance travelled by the pulse in travelling back to position `X = 2 m` (in going from `X` to left side) `+ 2 m` (in Velocity of propagation of the pulse `= (4 m)/(50 s) = (400 cm)/(50 s) = 8 cm//s`. |
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