#$&* course PHY 232 9:22 am 3/23 014. `Query 12
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Given Solution: `aSTUDENT SOLUTION WITH INSTRUCTOR COMMENTS: Since the rod is a rigid body it can form the smallest frequency corresponding to the largest wavelength only when held at the center. The smallest frequency corresponding to the largest wavelength is known as the fundamental. (a) the fingers act as a sort of pivot and the bar performs a teeter - totter motion on either side of the pivot **note: the motion is more like the flapping of a bird's wings, though it obviously doesn't result from the same sort of muscular action but rather from an elastic response to a disturbance, and the motion attenuates over time**. With this manner of motion, the ends have the greatest amount of amplitude, thus making them the antinodes. (b)Holding the rod at any point other than the center changes the wavelength of the first harmonic causing one end of the rod to have a smaller wave motion than the other. Since the fundamental frequency is in essence the smallest frequency corresponding to the largest wavelength, offsetting L would produce an inappropriatley sized wavelength. (c)Fundamental frequency of a steel rod: ( 5941 m /s) / [(2/1)(1.5m) = 1980.33Hz (d) v / (2/2) L = 3960.67Hz --achieved by holding the rod on one end. INSTRUCTOR COMMENTS: The ends of the rod are free so it can vibrate in any mode for which the ends are antinodes. The greatest possible wavelength is the one for which the ends are antinodes and the middle is a node (form ANA). Since any place the rod is held must be a node (your hand would quickly absorb the energy of the vibration if there was any wave-associated particle motion at that point) this ANA configuration is possible only if you hold the rod at the middle. Since a node-antinode distance corresponds to 1/4 wavelength the ANA configuration consists of 1/2 wavelength. So 1/2 wavelength is 1.5 m and the wavelength is 3 m. At propagation velocity 5941 m/s the 3 m wavelength implies a frequency of 5941 m/s / (3 m) = 1980 cycles/sec or 1980 Hz, approx.. The first overtone occurs with antinodes at the ends and node-antinode-node between, so the configuration is ANANA. This corresponds to 4 quarter-wavelengths, or a full wavelength. The resulting wavelength is 1.5 m and the frequency is about 3760 Hz. THE FOLLOWING INFORMATION MIGHT OR MIGHT NOT BE RELATED TO THIS PROBLEM: STUDENT RESPONSES WITH INSTRUCTOR COMMENTS INSERTED To find the amplitudes at 40, 20 and 10 cm from the left end: The amplitudes are: at 40 cm 0 at 20 cm .004m at 10 cm .002828 m I obtained my results by using the information in the problem to write the equation of the standing wave. Since the cosine function is maximum at 0, I substituted t=0 into the equation and the value of x that I wanted to find the amplitude for. ** wavelength = 192 m/2 / (240 Hz) = .8 m. Amplitude at x is .4 cm sin(2 `pi x / .8 m) **the maximum transverse velocity and acceleration at each of these points are found from the equation of motion: `omega = 2 `pi rad * 240 cycles / sec = 480 `pi rad/s. .004 m * 480 `pi rad/s = 6 m/s approx and .004 m * ( 480 `pi rad/s)^2 = 9000 m/s^2, approx. .0028 m * 480 `pi rad/s = 4 m/s approx and .0028 m * ( 480 `pi rad/s)^2 = 6432 m/s^2, approx. ** STUDENT COMMENT The last student response was very interesting. If I understand it correctly, she took the standing wave equation and found the max by setting the derivative equal to zero. INSTRUCTOR RESPONSE That is a very good statement of the reasoning illustrated in the student's solution.The time derivative of the position function is the velocity function for the particles of material, and if the particles at the point where you hold the bar isn't zero, energy will be lost at that point very quickly, causing the standing wave to dissipate its energy (or to retain too little energy to to form in the first place). Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!