Standing Waves Lab

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PHYSICS 23 LABORATORY E6: Standing Waves on a String NAME:

DATE:

Partner’s Name: Laboratory Instructor:

Laboratory Section:

Recitation Instructor:

Recitation Section:

OBJECTIVES: Reading: Introduction: This week in class you are beginning your study of wave motion, an example of which is light waves. In laboratory today we will study standing waves on a string with the ends fixed. (Incidentally, this is the mechanical analogue of a laser cavity.) Standing waves do not propagate through space (as traveling waves do). Consider transverse waves on a string that is clamped at both ends: in a traveling wave, a point on the string where the displacement y(x,t) has its maximum value will have an x-coordinate which increases or decreases as time goes on. But in a standing wave, the x-coordinate of the point never changes. A standing wave is formed when two traveling waves having the same frequency and speed, but traveling in opposite directions, combine according to the superposition principle, y total=y1+y2. Let y1(x,t) = A sin (kx-ωt) be a wave traveling in the +x direction and y2(x,t) = A sin (kx+ωt) be a wave traveling in the –x direction. Using the identity sin(A±B)=sinA cosB±cosA sinB, show that: y(x,t) = A sin (kxωt) + A sin (kx+ωt) = 2A sin ( kx ) cos ( ωt ).

Eq(1)

(As usual, k=2π/λ and ω=2πf, where λ is the wavelength and f is the frequency.) For a string of length L fixed at both ends, we need the amplitude to be 0 for all times at the endpoints, x = 0 and x = L. Show that this means λn = 2L/n, and hence a standing wave on the string has 1, 2, 3, ... half-wavelengths on the length L. A reminder that geometry requires fn λn = v, where v is the velocity of a traveling wave on a string. The velocity v = √(F/µ), where F is the tension in the string and µ is the mass density, mass per unit length. With F = Mng, the relation between the frequency and wavelength of the standing wave is fn λn = √( Mn g/µ ). Mn is the mass needed to provide the tension to cause the nth standing wave. Show that the mass Mn is given by Mn = fn2 λn2 µ /g = ( 4L2 fn2 µ ) / ( g n2 ). Apparatus: Sketch the apparatus.

Eq (2)

Procedure In this version of the experiment, we keep the tension in the string the same and vary the frequency of the vibrator. The mechanical vibrator should be as near one end of the string as possible. The string should just lightly fit into the clip on top of the vibrator. It should NOT press down hard on the vibrator. The amplitude (bottom knob on right) should be set near zero (nearly fully counterclockwise.) DO NOT exceed half scale (line on knob vertical.) Using a hanging mass of 200 grams or so, find the frequency of the first five harmonics, i.e. n=1,...,5. Arrange your results in a data table showing fn and the corresponding integer n. Measurements: Analysis: 1. What does Eq(2) imply about the relationship between fn and n? (Remember that Mn, L, µ and g are all constants in this experiment.) Graph f vs n. Determine µ (the mass per unit length of the string). Show how you calculated µ, being careful to use MKS units throughout. 2. How might you perform a check on your value for µ? . Conclusions: