Question

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of 343 m/s.

Answer 1

Answer: Length axis f= 114.3 Hz, Width axis f=228.67 Hz

Explanation:

We are given that,

Length of tub= 1.5 m

Width of tub= 0.75 m

Sound speed= 343 m/s

Now, we are also given shower is closed.

So, frequency is given as:

f= $m* \frac{v}{2L}$

For length axis

Put v= 343 m/s, m=1 and L=1.5 m

f= 1 * $\frac{343}{2*1.5}$

f= 114.3 Hz

For next resonant frequency, m=2

f= 2* $\frac{343}{2*1.5}$

f= 228.67

For width axis

Put v= 343 m/s, m=1 and L= 0.75 m

f= 1* $\frac{343}{2*0.75}$

f= 228.67 Hz

For next frequency, m=2

f= 2* $\frac{343}{2*0.75}$

f= 457.34 Hz