MHT CET · Physics · Waves and Sound
The equation of stationary wave on a string clamped at both ends and vibrating in third harmonic is \(Y=0.5 \sin (0.314 x) \cos (600 \pi t)\) where \(x\) and \(y\) are in \(\mathrm{cm}, t\) in second. The length of the vibrating string is
- A \(20 \mathrm{~cm}\)
- B \(10 \mathrm{~cm}\)
- C \(40 \mathrm{~cm}\)
- D \(30 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(30 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
The standard equation of a standing wave is
\(y=2 A \sin (k x) \cos (\omega t)\)
On comparing with standard equation,
\(\omega=600 \pi \mathrm{s}^{-1}\) and \(k=\frac{\pi}{10}\)
The wavelength of the wave can be obtained by using the wave number: \(k=\frac{2 \pi}{\lambda}=\frac{\pi}{10}\)
The wavelength is \(\lambda=20 \mathrm{~cm}\)
Frequency of the third harmonics is three times the fundamental.
Therefore, length of the string is: \(L=3\left(\frac{\lambda}{2}\right)=3(10)=30 \mathrm{~cm}\)
\(y=2 A \sin (k x) \cos (\omega t)\)
On comparing with standard equation,
\(\omega=600 \pi \mathrm{s}^{-1}\) and \(k=\frac{\pi}{10}\)
The wavelength of the wave can be obtained by using the wave number: \(k=\frac{2 \pi}{\lambda}=\frac{\pi}{10}\)
The wavelength is \(\lambda=20 \mathrm{~cm}\)
Frequency of the third harmonics is three times the fundamental.
Therefore, length of the string is: \(L=3\left(\frac{\lambda}{2}\right)=3(10)=30 \mathrm{~cm}\)
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