Question

A transverse wave produced near one end of an extremely long vibrating string is described by the equation below. (Ignore wave reflections from the other end of the string, which is extremely far ahead.) The linear density of this string is 0.0073 kg/m. What must be the tension applied to this string?

Answer 1

The tension that must be applied to this string = 477 N

Step-by-step explanation:

y(x,t)= 0.0321m sin (2.05x-524t + pi/4)

Comparing this to the general wave equation

y(x,t) = A sin (kx - wt + Φ)

where

A = amplitude of the wave = 0.0321 m

k = 2.05 /m

w = 524 /s

Φ = phase angle = pi/4

Velocity of a wave is given by

v = (w/k) = (524/2.05) = 255.61 m/s

Tension in the string is then related to the velocity of wave produced and the linear density through

v = √(T/μ)

v = velocity = 255.61 m/s

T = ?

μ = 0.0073 kg/m

v² = (T/μ)

T = v²μ = (255.61² × 0.0073) = 476.96 N

T = 477 N

Hope this Helps!!!