Wave Motion 2 Question 30
30. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y(x, t)=(0.01 m)\left[\sin \left(62.8 m^{-1}\right) x\right] \cos \left[\left(628 s^{-1}\right) t\right]$.
Assuming $\pi=3.14$, the correct statement(s) is (are)
(2012)
(a) the number of nodes is 5
(b) the length of the string is $0.25 m$
(c) the maximum displacement of the mid-point of the string from its equilibrium position is $0.01 m$
(d) the fundamental frequency is $100 Hz$
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Answer:
Correct Answer: 30. (b, c)
Solution:
Number of nodes $=6$
From the given equation, we can see that
$$ \begin{aligned} k & =\frac{2 \pi}{\lambda}=62.8 m^{-1} \\ \therefore \quad \lambda & =\frac{2 \pi}{62.8} m=0.1 m \\ l & =\frac{5 \lambda}{2}=0.25 m \end{aligned} $$
The mid-point of the string is $P$, an antinode $\therefore$ maximum displacement $=0.01 m$
$$ \begin{aligned} & \omega=2 \pi f & =628 s^{-1} \\ \therefore & f & =\frac{628}{2 \pi}=100 Hz \end{aligned} $$
But this is fifth harmonic frequency.
$\therefore$ Fundamental frequency $f _0=\frac{f}{5}=20 Hz$
