Question

The arrangement in the drawing shows a block (mass = 15.4 kg) that is held in position on a frictionless incline by a cord (length = 0.59 m). The mass per unit length of the cord is 1.18×10−21.18×10−2 kg/m, so the mass of the cord is negligible compared to the mass of the block. The cord is being vibrated at a frequency of 70.4 Hz (vibration source not shown in the drawing). What is the largest angle θ at which a standing wave exists on the cord?

Answer 1

Use the expression -

v = √T/µ [where v = speed of wave, T = tension in cord, µ = mass per unit length of the cord = 1.18 x 10^-2 kg/m]-------------(i)

Now-  

v = fλ

standing wave wavelength of lowest tension = λ = L/2 = 0.59 / 2 = 0.295 m

So -

v = 70.4 * 0.295 = 20.77 m/s

Put this in this expression (i) -

20.77 = √T/ (1.18 x 10^-2)

=> 431.31 = T/(1.18 x 10^-2)

=> T = 5.10 N

Again, we have -

T = mg(sin Θ) {vector component of mg acting parallel to incline}

=> 5.10 = 15.4(9.81)(sin Θ)

=> sin Θ = 5.10 / (15.4 x 9.81) = 0.0337

=> Θ = 1.93 deg. (Answer)