Question

The string shown in the figure below is driven at a frequency of 5.00 Hz. The amplitude of the motion is 15.0 cm and the wave speed is 19.5 m/s. Furthermore, the wave is such that y = 0 at x = 0 and t = 0.

(a) Determine the angular frequency for this wave.
? = rad/s
(b) Determine the wave number for this wave.
k = rad/m
(c) Write an expression for the wave function. (Use the following as necessary: x and t. Assume y is in cm, x is in m, and t is in s.)
y =
(d) Calculate the maximum transverse speed of an element of the string.
m/s
(e) Calculate the maximum transverse acceleration of an element of the string.
m/s2

Answer 1

a) Angular frequency can be found using the equation

\(\omega=2 \pi f\)

where \(f\) is frequency so we get

\(\omega=2 \pi(5.00 \mathrm{~Hz})=31.4 \mathrm{rad} / \mathrm{s}\)

b) To determine the wave number, \(k\), we first must find \(\lambda\). We can find \(\lambda\) by using the equation

\(\lambda=v / f\)

where \(v\) is velocity and \(f\) is frequency we have:

\(\lambda=19.5 / 5=3.9 \mathrm{~m}\)

and we can use this to find \(k\) using the equation

\(\mathrm{k}=2 \pi / \lambda\)

which is equal to

\(\mathrm{k}=2 \pi / 3.9 \mathrm{~m}=1.61 \mathrm{rad} / \mathrm{m}\)

c) just substitute the numbers you just found in for their values

\(u(x, t)=0.15 \sin \left(1.61 \frac{\mathrm{rad}}{m} x-31.4 \frac{\mathrm{rad}}{s} t\right)\)

in your case \(u(x, t)\) is \(y(x, t)\)

d) simply

\(v_{\max }=\omega A\)

which gives a value of

\(v_{\max }=\left(31.4 \frac{\mathrm{rad}}{x}\right)(0.15 \mathrm{~m})\)

which is \(4.71 \mathrm{~m} / \mathrm{s}\)

e) This is similar to problem \(\mathrm{d}\) ) except the equation is

\(a_{\max }=\omega^{2} A\)

plug in your numbers and you'll have an answer of

\(148 \mathrm{~m} / \mathrm{s}^{\wedge} 2\)