The equation of a wave on a string of linear mass density 0.02 kg m ⁻¹ is Y = 0.01sin [2π(t/0.02 - x/0.50)] m. The tension in the string is (a) 12.50 N (b) 6.25 N (c) 25 N (d) 50 N

Answer: (a). From the equation: Tperiod = 0.02 s ( f = 50 Hz), λ = 0.50 m v = fλ = 50 × 0.5 = 25 m/s Tension = μ v² = 0.02 × 625 = 12.5 N So the answer is (a) 12.50 N.

The equation of a wave on a string of linear mass density 0.02 kg m… | MHT CET 2026 (20 Apr, Shift 2)

Q24MHT CET 2026 · 20 APR · SHIFT 2Medium

Progressive Wave Equation

The equation of a wave on a string of linear mass density $0.02$ kg m $^{-1}$ is

$Y = 0.01\sin\left[2\pi\left(\frac{t}{0.02} - \frac{x}{0.50}\right)\right]$ m. The tension in the string is

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Answer: (a)

From the equation: $T_{period} = 0.02\text{ s}$ ( $f = 50\text{ Hz}$ ), $\lambda = 0.50\text{ m}$
$v = f\lambda = 50 \times 0.5 = 25\text{ m/s}$
Tension $= \mu v^2 = 0.02 \times 625 = 12.5\text{ N}$
So the answer is (a) 12.50 N.

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