Question

What is the frequency of a standing wave in a 1.5m closed pipe that is vibrating in the 5th harmonic if the speed of sound is 343m/s?
A
285Hz
B
390Hz
C
572Hz
D
620Hz

Answer 1

Simple Harmonic Motion
\[f=\frac{ \lambda }{ v }\]
there you go just input your value then you are good to go

Answer 2

@niyex The formula is wrong actually the correct formula is
\[v=f \lambda \]
then,
\[f=\frac{ v }{ \lambda }\]
but for the 5th harmonic motion we now that
\[\lambda=\frac{ 5v }{ 4L }\]
Read up on the harmonic motion of waves in a pipe closed at one end.
so we get our f to be
\[f=\frac{ 5v }{ 4L }\]
so we plug in the values we get
\[f=\frac{ 5(343) }{ 4(1.5) }\]
\[f=285.833 Hz\]
that would make your answer to be Option A

Answer 3

thanks @ijlal. but i found something strange in one of your equation \[\lambda=\frac{ 5v }{ 4L } =f \frac{ 5v }{ 4L }\] are you saying that \[\lambda=f\] if that what your insinuating then bro please check your equations well cos frequency is quite different from wavelength.
But i want to assume it is a mix up or a mistake

Answer 4

@niyex oh sorry it was a typo :D thanks for pointing it out ! the equation is correct the mistake is
\[\lambda = \frac{ 4L }{ 5 }\] so we put that in the equation
\[f=\frac{ v }{ \lambda }\]
it becomes
\[f=\frac{ 5v }{ 4L }\]