Question

A transverse wave is passing through a light string shown in the figure.The equation of wave is $y=A\sin (co\omega t-kx)$.The area of cross-section of string is $A$ and density is $p$.The hanging mass is:

• A. $A\omega$
• B. $\frac{\omega }{kg}$
• C. $\frac{\rho A{\omega }^2}{k^{2}g}$
• D. $\frac{k^{2}g}{\omega }$

Answer 1

The correct option is C $\frac{\rho A{\omega }^2}{k^{2}g}$

Force in the string = $mg$

Velocity of the wave=$\sqrt{\frac{T}{\mu }}=\sqrt{\frac{mg}{\rho A}}$ ----- (1)

Velocity of wave from the given equation = $\frac{\omega }{k}$ -------- (2)

When we equate both(1 and 2) the values we get : $m=\frac{{\omega }^2\rho A}{k^{2}g}$