a string 4.00 m long is held fixed At both ends. If a sharp blow is applied to t

Question

a string 4.00 m long is held fixed At both ends. If a sharp blow is applied to the string at its center, it takes 0.0300 s for the pulse to travel to the ends of the string and return to the middle. What is the fundamental frequency of oscillation for this string? a string 4.00 m long is held fixed At both ends. If a sharp blow is applied to the string at its center, it takes 0.0300 s for the pulse to travel to the ends of the string and return to the middle. What is the fundamental frequency of oscillation for this string? a string 4.00 m long is held fixed At both ends. If a sharp blow is applied to the string at its center, it takes 0.0300 s for the pulse to travel to the ends of the string and return to the middle. What is the fundamental frequency of oscillation for this string?

Explanation / Answer

F0 = fundamental Frequency
L = length of the tube
v = velocity of the sound wave
= wavelength
The fundamental frequency of a sound wave in a tube with either both ends OPEN or both ends CLOSED can be found using the following equation:
F0=V/2L
L can be found using the following equation:
L=/2
The wavelength, which is the distance in the medium between the beginning and end of a cycle, is found using the following equation:
=v/F0
This problem is semi-simple...
Step 1 Find v
Since its starting at the center and ending at the center, the length of the string is the distance traveled.

So d/t=v

or 4/0.0300=133.33 =v

So the tube or in this case string is 4m and L. This means
F0=V/2L

F0=133.33/[4(2)]

F0=16.66Hz

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