A pulse is sent traveling along a rope under a tension of 32 N whose mass per un

Question

A pulse is sent traveling along a rope under a tension of 32 N whose mass per unit length abruptly changes, from 15 kg/m to 37 kg/m. The length of the rope is 2.5 m for the first section and 2.9 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, t, between the two pulses detected at the origin?

Explanation / Answer

This problem is based on the simple relation that waves on a taut string travel with a speed given by

v=sqrt(T/mu), where "T" is the tension in the string and "mu" is its mass per unit length. So if v1 and v2 are the velocities of the pulse in the first and second sections respectively,

v1 = sqrt(32/15) = 1.46 m/s

v2 = sqrt(32/37) = 0.93 m/s.

Now that the speeds are known, it is just a matter of calculating travel times by noting the length of rope covered in each section. Until the discontinuity is first reached there is only one pulse, so we only need to consider the time elapsed after that instant, i.e. we can take that as the origin time.

Time taken by first reflected pulse to get back to the origin = t1 = l1/v1 = 2.5/1.46 = 1.7123 s

Time taken by the second pulse to travel from discontinuity to the wall and back (through the second section of rope) = 2 x l2/v2 = 2 x 2.9/0.93 = 6.2365 s

Time taken by second pulse to travel from the discontinuity back to the origin (through the first section of rope) = l1/v1 = t1.

Hence total time taken by the 2nd pulse, relative to origin time, is t2 = 6.2365 + t1

The required quantity is t2 - t1 = 6.2365 seconds.

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