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 ? 10?2 kg/m, creating a standing transverse wave. There is a 3.8-kg block hanging from the other end of the rope over a pulley. The oscillator has an angular frequency of 43.6 rad/s and an amplitude of 255.0 cm.

(a) What is the distance between adjacent nodes?

I thought it was 1.84266 but it was wrong...

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

I think it should be half of the (a)?

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

Shouldn't it be the hald of the (b)?

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

I solved it. It is zero!!

Explanation / Answer

a.

the velocity of the wave is,   

v = sqrt[mg/u] = sqrt[3.8*9.8/5.4 102] = 26.26 m/s

the wavelength of the oscillations of the oscillator is,

wavelength = v/f = v/[w/2*pi] = 2*pi*v/w = 2*pi*26.26 / 43.6 = 3.78 m

Thus, the distance between adjacent nodes is,

dx = wavelength/2 = 3.78/2 = 1.89 m

(b)

If the angular frequency of the oscillator doubles, the wavelength of the oscillations of the oscillator is,

wavelength = 2*pi*v/2w = 2*pi*26.26 /2*43.6 = 1.89 m

Thus, the distance between adjacent nodes is,

dx = wavelength/2 = 1.89/2 = 0.945 m

(c)

If the mass of the block is doubled, the velocity of the wave is,

  v = sqrt[mg/u] = sqrt[2*3.8*9.8/5.4 102] = 37.137 m/s

the wavelength of the oscillations of the oscillator is,

wavelength = v/f = v/[w/2*pi] = 2*pi*v/w = 2*pi*37.137 / 43.6 = 5.349 m

Thus, the distance between adjacent nodes is,

dx = wavelength/2 = 5.349/2 = 2.67 m

(d)

the amplitude is not affect on the wavelength of the wave. So, the chnage in distance is zero.

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