Question

DQuestion 5 1 pts A simple harmonic oscillator at the point x-0 generates a wave on a horizontal 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 stretched with a tension of 5.00 N. Find the maximum transverse acceleration of points on the rope, in m/s? Sample submission: 1230 Note: your answer should be much larger than g. which is why we are able to ignore gravity when deriving the wave speed. Submit Quiz No new data to save. Last checked at 8:44pm

Answer 1

Sorry friend,according to rules i have to submit first question answer only.

Given problem,frequency f = 40 Hz

amplitude A =3 cm =3 *10^-2

               A =0.03 m

Linear density u = 50 g/cm = 50 * 10^-3 kg/m

Tension T = 5 N

Maximum transverse acceleration a(max) =?

We know that displacement equation of the wave Y(x,t) = A cos (kx -wt + Ф )

At x =0 ,then Ф = 0 , then Y(x,t) =A cos (kx -wt)

acceleration a(y) = d^2 y /dt^2 = - Aw^2 cos (kx -wt)

therefore maximum transverse acceleration a(max) = Aw^2

But angular velocity , w = 2 π f

w = 2 * π * 40 =251.33 rad/s

a (max) =0.03 *(251.33)^2 = 0.03 * 63165.47 =1894.96 m/s^2.

a (max) ~ 1895 m/s^2.