MHT CET · Physics · Waves and Sound
A transverse wave travelling along a stretched string has a speed of \(30 \mathrm{~m} / \mathrm{s}\) and a frequency of 250 Hz. The phase difference between two points on the string 10 cm apart at the same instant is
- A \(0 \mathrm{ radian}\)
- B \(\left(\frac{\pi}{2}\right) \mathrm{ radian}\)
- C \(\left(\frac{5 \pi}{3}\right) \text{ radian}\)
- D \(\left(\frac{8 \pi}{3}\right) \text{ radian}\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{5 \pi}{3}\right) \text{ radian}\)
Step-by-step Solution
Detailed explanation
The wavelength of the wave is given by,
\(\lambda=\frac{\mathrm{v}}{\mathrm{n}}=\frac{30}{250}=0.12 \mathrm{~m}\)
Also, the phase difference ' \(\phi\) ' is given by, \(\phi=\frac{2 \pi}{\lambda} \times\) path difference
\(\begin{aligned}
& \therefore \quad \phi=\frac{2 \pi(0.1)}{\lambda} \\
& \text {...(path difference }= \\
& 10 \mathrm{~cm}=0.1 \mathrm{~m} \text { ) } \\
& \therefore \quad \phi=\frac{2 \pi}{0.12} \times 0.1 \\
& \therefore \quad \phi=\left(\frac{5 \pi}{3}\right)^{\text{radian}}
\end{aligned}\)
\(\lambda=\frac{\mathrm{v}}{\mathrm{n}}=\frac{30}{250}=0.12 \mathrm{~m}\)
Also, the phase difference ' \(\phi\) ' is given by, \(\phi=\frac{2 \pi}{\lambda} \times\) path difference
\(\begin{aligned}
& \therefore \quad \phi=\frac{2 \pi(0.1)}{\lambda} \\
& \text {...(path difference }= \\
& 10 \mathrm{~cm}=0.1 \mathrm{~m} \text { ) } \\
& \therefore \quad \phi=\frac{2 \pi}{0.12} \times 0.1 \\
& \therefore \quad \phi=\left(\frac{5 \pi}{3}\right)^{\text{radian}}
\end{aligned}\)
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