As in the figure below, a simple harmonic oscillator is attached to a rope of li

Question

As in the figure below, a simple harmonic oscillator is attached to a rope of linear mass density 5.4 102 kg/m, creating a standing transverse wave. There is a 4.1-kg block hanging from the other end of the rope over a pulley. The oscillator has an angular frequency of 44.7 rad/s and an amplitude of 252.0 cm.

(a) What is the distance between adjacent nodes?


(b) If the angular frequency of the oscillator doubles, what happens to the distance between adjacent nodes? (Enter the new distance.)


(c) If the mass of the block is doubled instead, what happens to the distance between adjacent nodes? (Enter the new distance.)


(d) If the amplitude of the oscillator is doubled, what happens to the distance between adjacent nodes? (Enter the change in distance.)

Explanation / Answer

here,

linear mass density , u = 5.4 * 10^-2 kg/m

tension in the string , T = m*g

T = 40.18 N

angular frequency , w = 44.77 rad/s

f = 44.77/2pi rad/s

amplitude , A = 252 cm

(a)

the distance between the two adjacenet nodes , l = v/f

l = sqrt(T/u) /(w/2pi)

l = sqrt( 40.18/0.054)/( 44.77/(2*pi))

l = 3.83 m

the distance between the two adjacenet nodes is 3.83 m

(b)

if w' = 2*w

using , the distance between the two adjacenet nodes , l = v/f

l = sqrt(T/u) /(w/2pi)

l' = l/2

l' = 1.91 m

the new distnace is 1.91 m

(c)

if the mass of block is doubled , m' = 2*m

the distance between the two adjacenet nodes , l = v/f

l = sqrt(mg/u) /(w/2pi)

then, l' = sqrt(2) * l

l' = 1.41*l

l' = 5.4 m

the new distance is 5.4 m

(d)

there is no change in the wavelength

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