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, 0.75m wide and 1.5m 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. a) what are the lowest two frequencies that correspond to resonances on the short axis? b) what are the lowest two frequencies that correspond to resonances on the longer axis?

Explanation:

For first overtone

Standing waves will be formed lengthwise and breadth-wise in the enclosures  having dimension of .75m  x 1.5 m

A ) For the formation of lowest two frequencies formed by standing waves along the breadth  , fundamental note and first overtone may be considered.  

For fundamental note ,  

the condition is  

wave length λ = 2L = 2 x 0.75 m  

λ = 1.5 m  

frequency n = v / λ

= 343 / 1.5  

= 229  Hz approx

For first overtone

λ = L = 0.75m

frequency n = v / λ

n = 343 / 0.75  

= 457 Hz approx

B)

For the formation of lowest two frequencies formed by standing waves along the length , fundamental note and first overtone may be considered.  

For fundamental note ,  

the condition is  

wave length λ = 2L = 2 x 1.5 m  

λ = 3 m  

frequency n = v / λ

= 343 / 3  

= 114 Hz approx

frequency n = v / λ

n = 343 / 1.5  

= 229 Hz approx

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

report!

explanation:

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