A. A string has a linear density of 7.6 x 10 -3 kg/m and is under a tension of 3
ID: 1332340 • Letter: A
Question
A. A string has a linear density of 7.6 x 10-3 kg/m and is under a tension of 380 N. The string is 2.3 m long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that make up the standing wave.
B. A tube with a cap on one end, but open at the other end, has a fundamental frequency of 134.5 Hz. The speed of sound is 343 m/s. (a) If the cap is removed, what is the new fundamental frequency of the tube? (b) How long is the tube?
Explanation / Answer
A) A standing wave is composed of two oppositely traveling waves. The speed v of these waves is given by
v= / F/ (m/L)
where F is the tension in the string and
m/L is its linear density (mass per unit length). Both F and m/L are given in the statement of the problem.
The wavelength of the waves can be obtained by visually inspecting the standing wave pattern.
The frequency of the waves is related to the speed of the waves and their wavelength by f = v/
a) The speed of the waves is v= / 380/ (7.6 x 10-3) = 223.61 m/s
b) Two loops of any standing wave comprise one wavelength. Since the string is 2.3 m long and consists of three loops (see the drawing), the wavelength is = 2/3(2.3) = 1.533 m
c. The frequency of the waves is f= v/= 223.61/1.533= 145.86 Hz
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B) Given f1oc = 134.5 Hz, oc is open close , oo is open open (say)
n=1, v=343 m/s
fnoc = n(v/4L) and fn00 = n(v/2L)
a) We don’t need to know v or L, since they are the same in both cases. Solve each equation for v/L and set equal
v/L = 4f1oc and v/L = 2 f1oo means f1oo = 2f1oc = 2(134.5) =269 Hz
b) Can solve for L from either open-open or open-closed tubes
f1oc=1(v/4L) means L = v/(4f1oc) =343/(4x134.5) =0.6375 m
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