Wave Motion 2 Question 3

3. A string is clamped at both the ends and it is vibrating in its 4 th harmonic. The equation of the stationary wave is $Y=0.3 \sin (0.157 x) \cos (200 \pi t)$. The length of the string is (All quantities are in SI units)

(Main 2019, 9 April I)

(a) $60 m$

(b) $40 m$

(c) $80 m$

(d) $20 m$

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Answer:

Correct Answer: 3. (c)

Solution:

  1. Given equation of stationary wave is

$$ Y=0.3 \sin (0.157 x) \cos (200 \pi t) $$

Comparing it with general equation of stationary wave, i.e. $Y=a \sin k x \cos \omega t$, we get

$$ \begin{aligned} & k & =\left(\frac{2 \pi}{\lambda}\right)=0.157 \\ \Rightarrow & \lambda & =\frac{2 \pi}{0.157}=4 \pi^{2} \quad\left(\because \frac{1}{2 \pi} \approx 0.157\right) . \\ \text { and } & \omega & =200 \pi=\frac{2 \pi}{T} \Rightarrow T=\frac{1}{20} s \end{aligned} $$

As the possible wavelength associated with $n$th harmonic of a vibrating string, i.e. fixed at both ends is given as

$$ \lambda=\frac{2 l}{n} \quad \text { or } \quad l=n\left(\frac{\lambda}{2}\right) $$

Now, according to question, string is fixed from both ends and oscillates in 4th harmonic, so

$$ 4\left(\frac{\lambda}{2}\right)=l \Rightarrow 2 \lambda=l $$

or $l=2 \times 4 \pi^{2}=8 \pi^{2} \quad$ [using Eq. (i)]

Now, $\quad \pi^{2} \approx 10 \Rightarrow l \approx 80 m$



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