The equation of a wave on a string of linear mass density 0.04 kg m⁻¹ is given by y = 0.02 (m)sin [2π((t)/(0.04 (s)) - (x)/(0.50 (m)))]. The tension in the string is (a) 4.0 N (b) 12.5 N (c) 0.5 N (d) 6.25 N

Answer: (d). From the equation: period 0.04 s ( f = 25 Hz), λ = 0.50 m v = fλ = 25 × 0.5 = 12.5 m/s Tension = μ v² = 0.04 × 156.25 = 6.25 N So the answer is (d) 6.25 N.

The equation of a wave on a string of linear mass density 0.04 kg m⁻¹… | JEE Main 2010 (AIEEE)

Q92JEE Main 2010 · AIEEEMedium

Progressive Wave Equation

The equation of a wave on a string of linear mass density $0.04\text{ kg m}^{-1}$ is given by $y = 0.02\,(\text{m})\sin\left[2\pi\left(\frac{t}{0.04\,(\text{s})} - \frac{x}{0.50\,(\text{m})}\right)\right].$

The tension in the string is

Show answer and solution

Answer: (d)

From the equation: period $0.04\text{ s}$ ( $f = 25\text{ Hz}$ ), $\lambda = 0.50\text{ m}$
$v = f\lambda = 25 \times 0.5 = 12.5\text{ m/s}$
Tension $= \mu v^2 = 0.04 \times 156.25 = 6.25\text{ N}$
So the answer is (d) 6.25 N.

Report a mistake

Progressive Wave Equation question from JEE Main 2010 (AIEEE)

More questions on Progressive Wave Equation