Question

The arrangement in the drawing shows a block (mass = 14.6 kg) that is held in position on a frictionless incline by a cord (length = 0.599 m). The mass per unit length of the cord is 1.20

Answer 1

T = m g sin(theta)

where m is the mass of the block

c = Sqrt[T/u]

where T is the tension and u is the mass density of the string

u = 0.012

the fundamental mode as

f = c/2L where L is the length of the string

the allowable frequencies are integral multiples of the fundamental, so we have

f = n c/2L

c = 2 L f/n

2 *L* f/n = Sqrt[ m* g* sin(theta)/u]

sin(theta) = 4 L^2 f^2 u/(m g n^2)

L = 0.599      m = 14.6 kg        f = 167 Hz         u = 0.012     g =9.8

sin(theta) = 4*0.599^2*167^2*0.012/(14.6*9.8*n^2) = 3.3569 / n^2

now put the intger values of n

n can' be 1 because sin fun. can't be greater than 1

n = 2 , sin(theta) = 3.3569/4 = 57.05 deg

try n = 3 , sin(theta) = 3.3569/9 = 21.90 deg

n=4 , sin(theta) = 3.3569/16 = 12.11 can't because according to question angle should be greater than 15 deg.

so ans is 21.90 deg when n = 3