Question:
When two sound waves travel in the same direction in a medium, the displacements of a particle located at \(x\) at time \(t\) is given by :
\(y_1 = 0.05\cos(0.50\pi x - 100\pi t)\)
\(y_2 = 0.05\cos(0.46\pi x - 92\pi t)\)
where \(y_1\), \(y_2\) and \(x\) are in meters and \(t\) in seconds. The speed of sound in the medium is :
When two sound waves travel in the same direction in a medium, the displacements of a particle located at \(x\) at time \(t\) is given by :
\(y_1 = 0.05\cos(0.50\pi x - 100\pi t)\)
\(y_2 = 0.05\cos(0.46\pi x - 92\pi t)\)
where \(y_1\), \(y_2\) and \(x\) are in meters and \(t\) in seconds. The speed of sound in the medium is :
\(y_1 = 0.05\cos(0.50\pi x - 100\pi t)\)
\(y_2 = 0.05\cos(0.46\pi x - 92\pi t)\)
where \(y_1\), \(y_2\) and \(x\) are in meters and \(t\) in seconds. The speed of sound in the medium is :
Show Hint
Compare with \(y = A\cos(kx - \omega t)\) and use \(v = \omega/k\).
Updated On: Oct 31, 2026
- 92 m/s
- 200 m/s
- 100 m/s
- 332 m/s
Show Solution
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The Correct Option is B
Solution and Explanation
Compare with \(y = A\cos(kx - \omega t)\).
Wave 1: \(k_1 = 0.50\pi\), \(\omega_1 = 100\pi\), so \(v = \frac{\omega_1}{k_1} = \frac{100\pi}{0.50\pi} = 200\) m/s.
Wave 2: \(k_2 = 0.46\pi\), \(\omega_2 = 92\pi\), so \(v = \frac{92\pi}{0.46\pi} = 200\) m/s.
Both waves travel at the same speed, the speed of sound in the medium: 200 m/s.
Answer: (2) 200 m/s
Wave 1: \(k_1 = 0.50\pi\), \(\omega_1 = 100\pi\), so \(v = \frac{\omega_1}{k_1} = \frac{100\pi}{0.50\pi} = 200\) m/s.
Wave 2: \(k_2 = 0.46\pi\), \(\omega_2 = 92\pi\), so \(v = \frac{92\pi}{0.46\pi} = 200\) m/s.
Both waves travel at the same speed, the speed of sound in the medium: 200 m/s.
Answer: (2) 200 m/s
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