The distance between two consecutive points with phase difference of 45° in a wave of frequency 300 Hz is 4.0 m. The velocity of the travelling wave is (in km/s) (a) 1.6 (b) 3.6 (c) 4.8 (d) 9.6

Answer: (d). 45° = π/4 corresponds to Δ x = (λ)/(8), so λ = 8 × 4 = 32 m. v = fλ = 300 × 32 = 9600 m/s So the answer is (d) 9.6 km/s.

The distance between two consecutive points with phase difference of… | MHT CET 2025 (5 May, Shift 2)

Q30MHT CET 2025 · 5 MAY · SHIFT 2Easy

Progressive Wave Equation

The distance between two consecutive points with phase difference of $45^\circ$ in a wave of frequency 300 Hz is 4.0 m. The velocity of the travelling wave is (in $\text{km/s}$ )

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Answer: (d)

$45^\circ = \frac{\pi}{4}$ corresponds to $\Delta x = \frac{\lambda}{8}$ , so $\lambda = 8 \times 4 = 32$ m.
$v = f\lambda = 300 \times 32 = 9600$ m/s
So the answer is (d) $9.6$ km/s.

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