Question

A simple harmonic oscillator at the position x = 0 generates a wave on a string. The oscillator moves up and down at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. At time t = 0, the oscillator is passing through the origin and moving down. The string has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N.

Question 2 9 pts Consider the piece of string at x = 0.375 cm. g) What is its position at t = 2.5 seconds? Justify your answer with either math or words. h) What is its velocity (magnitude and direction) at t = 2.5 seconds? i) What is its acceleration at t = 2.5 seconds? j) What physical force acts on this piece of string to create its oscillating motion?

Answer 1

given
f = 40.Hz
angular frequency, w = 2*pi*f

= 2*pi*40

= 251 rad/s

Amplitude, A = 3.00 cm
mue = 50 g/m = 0.050 kg/m
T = 5 N

wave speed on the string, v = sqrt(T/mue)

= sqrt(5/0.05)

= 10 m/s

wave number, k = w/v

= 251/10

= 25.1 rad/m

now use,

y(x,t) = A*sin(k*x - w*t)

at x = 0.375 cm
g) at t = 2.5 s

y(x,t) = 3*sin(25.1*0.375 - 251*2.5)

= -2.17 cm

h) v = dy/dt

= -A*w*cos(k*x - w*t)

= -3*251*cos(25.1*0.375 - 251*2.5)

= 521 cm/s^2 (or) 5.21 m/s^2

i) a = -dv/dt

= -A*w^2*cos(k*x - w*t)

= -3*251^2*sin(25.1*0.375 - 251*2.5)

= 1.36*10^5 cm/s^2 (or) 1.36*10^3 m/s^2

j) The tension in the string ats on this piece of string to creat its oscillating motion.