Question:
The pressure wave, \(P = 0.01\sin[1000t - 3x]\ \mathrm{Nm^{-2}}\), corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is \(0^\circ\mathrm{C}\). On some other day, when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be \(336\ \mathrm{ms^{-1}}\). Approximate value of T is
The pressure wave, \(P = 0.01\sin[1000t - 3x]\ \mathrm{Nm^{-2}}\), corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is \(0^\circ\mathrm{C}\). On some other day, when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be \(336\ \mathrm{ms^{-1}}\). Approximate value of T is
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Read the speed at \(0^\circ\mathrm{C}\) as \(\omega/k\), then use \(v \propto \sqrt{T}\) (T in kelvin).
Updated On: Nov 5, 2026
- \(12^\circ\mathrm{C}\)
- \(15^\circ\mathrm{C}\)
- \(4^\circ\mathrm{C}\)
- \(11^\circ\mathrm{C}\)
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The Correct Option is C
Solution and Explanation
From the wave equation, \(\omega = 1000\) and \(k = 3\), so \(v_0 = \frac{\omega}{k} = \frac{1000}{3} = 333.3\) m/s at 273 K.
\(v \propto \sqrt{T}\): \(\frac{T}{273} = \left(\frac{336}{333.3}\right)^2 = (1.008)^2 = 1.016\).
\(T = 273 \times 1.016 \approx 277.4\) K \(\approx 4.4^\circ\mathrm{C}\), i.e. about \(4^\circ\mathrm{C}\).
Answer: (3) \(4^\circ\mathrm{C}\)
\(v \propto \sqrt{T}\): \(\frac{T}{273} = \left(\frac{336}{333.3}\right)^2 = (1.008)^2 = 1.016\).
\(T = 273 \times 1.016 \approx 277.4\) K \(\approx 4.4^\circ\mathrm{C}\), i.e. about \(4^\circ\mathrm{C}\).
Answer: (3) \(4^\circ\mathrm{C}\)
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