Wave Motion Ques 40
- 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: 40.(c)
Solution:
Formula:
- 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, $\quad \pi^{2} \approx 10 \Rightarrow l \approx 80 m$
BETA
