1. Wire A is stretched tightly between two posts. The length of wire A is 1 m an
ID: 1294365 • Letter: 1
Question
1. Wire A is stretched tightly between two posts. The length of wire A is 1 m and it vibrates in its fundamental mode with a frequency of 10 Hz. Wire B is made of the same material as A and is under the same tension as A, but is stretched between two different posts. It is observed that the frequency of the second harmonic of wire B is 10 Hz. This tells you that the length of wire B must be 2.0 meters True or False?
2. A 0.40 m string is clamped at both ends. The lowest frequency for a standing wave is 325 Hz. From this, you conclude that the wave speed in the string is equal to
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Explanation / Answer
1) frequency= nv/2L for the first string 2L= nv/f, v are contant because tension so for the first string 1/10=2L and for the second string 2/10=2L so TRUE because value for L doubles
2)325= 1v/2(.40) s0 v= 325*2*.40/1= 260
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