Question:

A string of length \(0.5\) m and mass \(10^{-3}\) kg is tightly clamped at its ends. The tension in the string is \(0.8\) N. Identical wave pulses are produced at one end at equal intervals of time \(\Delta t\). What is the minimum value of \(\Delta t\) which allows constructive interference between successive pulses?

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Find the wave speed, then the round trip time between the clamped ends.
Updated On: Oct 1, 2026
  • \(0.40\) s
  • \(0.20\) s
  • \(0.10\) s
  • \(0.05\) s
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The Correct Option is D

Solution and Explanation

Step 1: Wave speed
Linear density \(\mu=\frac{10^{-3}}{0.5}=2\times10^{-3}\) kg/m. \(v=\sqrt{\frac T\mu}=\sqrt{\frac{0.8}{2\times10^{-3}}}=20\) m/s.

Step 2: Round trip
A pulse reflects at a clamped end with a phase change of \(\pi\). After reflecting at both ends it has two phase changes, so it returns in its original form after \(\frac{2L}{v}\).
\(\frac{2L}{v}=\frac{2\times0.5}{20}=0.05\) s.

Step 3: Result
New pulses must be produced every \(0.05\) s so that they add constructively with the returning pulse. Option (D).

Final Answer:
The minimum interval is \(0.05\) s, option (D). \[ \boxed{\text{(D) }0.05\text{ s}} \]
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