Question

A simple harmonic oscillator at the point x=0 genrates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. the rope has a linear mass density of `50.0 g//m` and is strectches with a tension of 5.00 N. (a) determine the speed of the wave. (b) find the wavelength. (c ) write the wave function y(x,t) for the wave, Assume that the oscillator has its maximum upward displacement at time t=0. (d) find the maximum transverse acceleration at time t=0. (d) find the maximum transverse of points on the rope. (e) in the discussion of transverse waves in this chapter, the force of gravity was ignored. is that a reasonable assumption for this wave? explain.

Answer 1

(a). `V=sqrtF//mu=sqrt((5.00 N)//(0.0500 kg//m))=10.0 m//s`
(b). `Lambda= c//f=)10.0 m//s)//(40.0 Hz)=0.250 m`
(c ) `y(x,t)=A cos(kx-omegat) ("note":y(0.0)=A`, as specified.)`
`k=2pi//lambda=8pi rad//m, omega=2pif=80.0pi rad//s`
`y(x,t)=(300 cm)cos[pi(8.00 rad//m)x-(80.0pi rad//s)t]`
(d). `v_(y)=+Aomegasin(kx-omegat) and a_(y)=-Aomega^(2)cos(kx-omegat)`
`a_(y,max)=Aomega^(2)=A(2pif)^(2)=1890 m//s^(2)`
(e) `a_(y,max)` is much larger than `g`, so `g` can be ignored.