12a: Two identical strings (same mass and length) are hung from the ceiling. Str

Question

12a: Two identical strings (same mass and length) are hung from the ceiling. String 1 supports a mass M while string 2 supports a mass 2M. The masses are shifted back and forth horizontally which results in transverse waves propagating up the strings. Assume that M » Mstring. On which string is the wave propagation speed larger? By what factor? Explain b: Assume that the masses are shifted back and forth at the same frequency on each string On which string is the wavelength longer? By what factor? Explain. 13a: A string on a guitar is oscillating at 200 Hz as shown (previous page). The equilibrium length of the string is 0.5 m and its mass is 0.0025 kg. Determine the tension in the string b: Determine the fundamental frequency of this string : What would be the frequency if the string were oscillating with six nodes (not counting the node at each end of the string)? Use the tension from part a d: The string tension is changed to 50 N and the frequency is still 200 Hz. Is there a standing wave on the string now? If the answer is yes, then draw it. If the answer is no, then explain why not. e: The string is currently at the position labeled A in the diagram. Assume that that position represents maximum displacement for the string. Draw and explain the appearance of the string exactly one-fourth of an oscillation cycle later. 14: A guitar string of length L has a fundamental frequency of 150 Hz. The person playing the guitar places their finger at x = L/3 so that there must be a node there now. He then plucks the string between 2-0 and x = L/3. What is the new fundamental frequency?

Explanation / Answer

12.(A) v = sqrt[ T / linear mass densiy]

both are same stringso linear mass density will be same.

T1 = M g and T2 = 2 M g

so propagation speed will be larger in 2M hanging string.

(b) wavelength = speed/ frequency.

with 2 M .

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