Standing waves on a 1.0 m long string that is fixed at both ends are seen at successive
frequencies of 36 Hz and 48 Hz.
(a) What are the fundamental frequency and the wave speed?
(b) Draw the standing wave pattern when the string oscillates at 60 Hz. Which mode of
vibration does this correspond to?
(a) To find the fundamental frequency and wave speed, we can use the formula:
f = nv/2L
where f is the frequency, n is the harmonic number, v is the wave speed, and L is the length of the string.
For the first standing wave with a frequency of 36 Hz, we can set n = 1:
36 = (1)(v)/(2)(1.0)
v = (36)(2)(1.0) = 72 m/s
For the second standing wave with a frequency of 48 Hz, we can set n = 2:
48 = (2)(v)/(2)(1.0)
v = (48)(2)(1.0) = 96 m/s
So, the wave speed is 72 m/s and 96 m/s.
To find the fundamental frequency, we can set n = 1 and use either wave speed:
f = (1)(72)/(2)(1.0) = 36 Hz
f = (1)(96)/(2)(1.0) = 48 Hz
So, the fundamental frequency is 36 Hz.
(b) When the string oscillates at 60 Hz, we need to first determine the harmonic number n. We can rearrange the formula to solve for n:
n = 2Lf/v
Plugging in the values, we have:
n = (2)(1.0)(60)/(72) = 1.67
Since harmonic numbers must be whole numbers, we can round up or down to the nearest whole number. In this case, n is approximately 2.
The standing wave pattern for the second mode of vibration (n = 2) is called the first overtone. It looks like a single full wave with two nodes.