Physics - Wave on String Question with Solution | TestHub

The linear mass density of the string is given by:

$\mu =\frac{m}{L}=\frac{10^{-3}}{0.5}=2\times 10^{-3}\text{ kg/m}$

The speed of the transverse wave pulse on the string is:

$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{0.8}{2\times 10^{-3}}}=\sqrt{400}=20\text{ m/s}$

When a wave pulse is produced at one end, it travels to the other end, reflects, and comes back. Since the string is tightly clamped at both ends, the pulse undergoes a phase change of $\pi$ (inversion) upon reflection at the far end, and another phase change of $\pi$ upon reflection at the near end.

After two reflections (one complete round trip), the total phase change is $2\pi$. This means the returning pulse is upright and traveling in the original direction, exactly matching its initial state.

For constructive interference to occur between successive pulses, the second pulse must be produced exactly when the first pulse has completed one round trip and is starting its second trip.

The minimum time interval $\Delta t$ is the time taken for one round trip:

$\Delta t=\frac{2L}{v}=\frac{2\times 0.5}{20}=\frac{1}{20}=0.05\text{ s}$

Answer: $0.05$ s