Equation of travelling wave on a stretched string of linear density $5 \mathrm{~g} / \mathrm{m}$ is $y=0.03 \sin (450 \mathrm{t}-9 \mathrm{x})$ where distance and time are measured in SI united. The tension in the string is:

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Waves & Sound

Equation of travelling wave on a stretched string of linear density

5 g / m

y = 0.03 sin (450 t - 9 x)

where distance and time are measured in SI united. The tension in the string is:

Answer: (C)

Compare

y = 0.03 sin (450 t - 9 x)

with

y = A sin (ω t - k x)

to read off

ω = 450 rad/s

and

k = 9 rad/m

, so the wave speed is

v = \frac{ω}{k} = \frac{450}{9} = 50 m/s

. On a stretched string

v = T / μ

, and with linear density

μ = 5 g/m = 5 \times 1 0^{- 3} kg/m

the tension is

T = μ v^{2} = 5 \times 1 0^{- 3} \times 5 0^{2} = 12.5 N

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option