MHT CET · Physics · Waves and Sound
For a stationary wave, \(Y=10 \sin \left(\frac{\pi x}{15}\right) \cos (48 \pi t) \mathrm{cm}\), the distance between a node and the successive antinode is
- A \(7.5 \mathrm{~cm}\)
- B \(30 \mathrm{~cm}\)
- C \(15 \mathrm{~cm}\)
- D \(60 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(7.5 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Comparing given equation with standard equation \(y=2 a \sin \left(\frac{2 \pi x}{\lambda}\right) \cos (2 \pi f t)\)
where, \(a\) is the amplitude, \(k=\frac{2 \pi}{\lambda}\) the propagation constant and \(f\) the frequency.
Therefore, we obtain: \(\frac{2 \pi}{\lambda}=\frac{\pi}{15} \mathrm{~cm}^{-1}\)
\(\Rightarrow \lambda=30 \mathrm{~cm}\)
Distance between nearest node and antinode \(=\frac{\lambda}{4}=\frac{30}{4}=7.5 \mathrm{~cm}\)
where, \(a\) is the amplitude, \(k=\frac{2 \pi}{\lambda}\) the propagation constant and \(f\) the frequency.
Therefore, we obtain: \(\frac{2 \pi}{\lambda}=\frac{\pi}{15} \mathrm{~cm}^{-1}\)
\(\Rightarrow \lambda=30 \mathrm{~cm}\)
Distance between nearest node and antinode \(=\frac{\lambda}{4}=\frac{30}{4}=7.5 \mathrm{~cm}\)
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