Wave Motion 5 Question 5
5. A string of length $0.4 m$ and mass $10^{-2} kg$ is tightly clamped at its ends. The tension in the string is $1.6 N$. Identical wave pulses are produced at one end at equal intervals of time $\Delta t$. The minimum value of $\Delta t$, which allows constructive interference between successive pulses, is
(1998, 2M)
(a) $0.05 s$
(b) $0.10 s$
(c) $0.20 s$
(d) $0.40 s$
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Answer:
Correct Answer: 5. (b)
Solution:
- Mass per unit length of the string,
$$ m=\frac{10^{-2}}{0.4}=2.5 \times 10^{-2} kg / m $$
$\therefore$ Velocity of wave in the string,
$$ \begin{aligned} & v=\sqrt{\frac{T}{m}}=\sqrt{\frac{1.6}{2.5 \times 10^{-2}}} \\ & v=8 m / s \end{aligned} $$
For constructive interference between successive pulses
$$ \begin{aligned} \Delta t _{\text {min }} & =\frac{2 l}{v} \\ & =\frac{(2)(0.4)}{8}=0.10 s \end{aligned} $$
(After two reflections, the wave pulse is in same phase as it was produced, since in one reflection its phase changes by $\pi$, and if at this moment next identical pulse is produced, then constructive interference will be obtained.)
