Vertical Oscillation: Solving Wave Problems with Frequency, Amplitude, and Mass

[starhallie]

Homework Statement

A wire is 2m in length with a mass of 56g. One end is attached to a support, and the other end is attached to a hanging mass of 5kg. The support oscillates vertically with a frequency of 150Hz and 3cm amplitude.
a) What is the speed of a wave as it travels along the string?
b) What is the wavelength of the traveling wave?
c) What is the maximum transverse displacement of the string?
d) What is the maximum transverse speed of the string? Where does it occur?
e) What is the maximum transverse acceleration of the string? Where does it occur?

Homework Equations

The only equations I'm sure about are:
a) v= wavelenth/T= (f)(wavelength)
b) wavelength= v/f

The Attempt at a Solution

I have a lot of trouble with wave problems and I missed one of the lectures for my class on waves, and you guys are just so helpful here. I was hoping someone could get me started in the right direction on this problem, and I'd really appreciate the help!

 

[jbsteele]

a) Speed of a wave as it travels along the string: v= (150Hz)(wavelength)b) Wavelength of the traveling wave: wavelength= v/f = v/(150Hz)c) Maximum transverse displacement of the string: d) Maximum transverse speed of the string: Where does it occur?e) Maximum transverse acceleration of the string: Where does it occur?

 

[nlightner]

First, let's define some variables. The length of the wire, L, is 2m and its mass, m, is 56g. The hanging mass, M, is 5kg. The frequency of oscillation, f, is 150Hz and the amplitude, A, is 3cm. We can also define the speed of the wave, v, as well as the wavelength, λ, maximum transverse displacement, ymax, maximum transverse speed, vmax, and maximum transverse acceleration, amax.

a) To find the speed of the wave, we can use the equation v = fλ. Plugging in the given values, we get v = (150Hz)(λ). We need to convert the amplitude from centimeters to meters, so A = 0.03m. Thus, we can rearrange the equation to solve for the wavelength: λ = v/f = (0.03m)/(150Hz) = 0.0002m/s. So, the speed of the wave is 0.0002m/s.

b) Now that we know the speed of the wave, we can use the equation λ = v/f to find the wavelength. Plugging in the values, we get λ = (0.0002m/s)/(150Hz) = 0.000001333m. Therefore, the wavelength of the wave is 0.000001333m.

c) To find the maximum transverse displacement, we can use the equation ymax = A = 0.03m. This means that the maximum transverse displacement is 3cm.

d) To find the maximum transverse speed, we can use the equation vmax = 2πfA. Plugging in the values, we get vmax = (2π)(150Hz)(0.03m) = 28.274m/s. This maximum transverse speed occurs when the string is at its maximum displacement, which is 3cm from the equilibrium position.

e) To find the maximum transverse acceleration, we can use the equation amax = (2πf)^2A. Plugging in the values, we get amax = (2π(150Hz))^2(0.03m) = 14137.155m/s^2. This maximum transverse acceleration occurs when the string is at its maximum displacement, which is 3cm from the equilibrium position.