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Superposition of Waves — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsSuperposition of Waves1 Mark
MCQ
A pipe closed at one end has length $0.8 m$. At its open end a $0.5 m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in $50 N$ and the speed of sound is $320 m / s$, the mass of the string used is
  • A
    4 gram
  • ✓
    10 gram
  • C
    8 gram
  • D
    12 gram
✓

Answer

Correct option: B.
10 gram
(b) : Let, length of string, $l=0.5 m$
Length of pipe, $l=0.8 m$
As the frequency of string is vibrating in second harmonics i.e., $v=\left(\frac{2}{2 l}\right) \sqrt{\frac{T}{\mu}} \quad(\mu=$ mass pen unit length of string) of string)
Fundamental frequency of closed pipe is, $v^{\prime}=\frac{v}{4 L}$. Here ' $v$ ' is speed of sound.
At resonance, $v=v^{\prime}$
$
\begin{aligned}
& \Rightarrow\left(\frac{2}{2 l}\right) \sqrt{\frac{T}{\mu}}=\frac{v}{4 L} \\
& \Rightarrow\left(\frac{1}{0.5}\right) \sqrt{\frac{50}{\mu}}=\left(\frac{320}{4 \times 0.8}\right) \Rightarrow \frac{50}{\mu}=2500 \\
& \Rightarrow \mu=\frac{1}{50} ; \text { mass of string, } m=\mu l=\frac{1}{50} \times 0.5 \\
& =\left(\frac{1}{100}\right) kg =10 gm
\end{aligned}
$

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