. If a string fixed at both ends resonates in its fundamental mode with a frequency...

. If a string fixed at both ends resonates in its fundamental mode with a frequency of 150 Hz, at which of the following frequencies will it not resonate?

Expert Solution

A wave traveling in one direction on the string reflects off the end, and returns inverted because the end is fixed. This gives two identical waves traveling in opposite directions on the string, just what is needed for a standing wave.

Constructive interference takes place for all reflected waves simultaneously only when the wavelength is related to the length L of the string by:

n(lambda)/2

= L, where n = 1, 2, 3, ...

Using f = v/lambda the corresponding frequencies are:

f_n

=

nv/2L

, where n = 1, 2, 3, ...

A large-amplitude standing wave builds up because of resonance for these special frequencies only. Note that there is always a node at a fixed end.

The lowest resonance frequency (n=1) is known as the fundamental frequency for the string. All the higher frequencies are known as harmonics - these are integer multiples of the fundamental frequency.

so only at integer multiples of 150 Hz will the string resonate ex. 150, 300, 450, 600 and so on.

At any other frequency it will not resonate

genius_generous answered 1 year ago

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