Solving the Frequency of Sound Waves at 20°C

[samisoccer9]

1. a source of sound is directed a large brick wall. a girl walks towards the wall and notices that the intensity of the sound decreases to aminimum every 50cm. what is the frequency of the sound? the air temperature is 20 degrees celsius (answer is 344Hz)2. attempted to use the universal wave equation, being V = frequency x lambda3. found new velocity of sound to be 343.2m/s by doing 331.4 m/s (0.59)(20degrees celsius)
took this new velocity, and attempted to put it into the above listed equation in #2, although the equation is meant for the velocity of a WAVE, i did not know which equation to use. help please and thanks !?

 

[NascentOxygen]

Hi samisoccer9. Have you drawn a diagram showing the incident and reflected waves? Please attach it here.

BTW, the title of your post would be better as INTERFERENCE OF WAVES, not RESONANCE.

 

[samisoccer9]

I haven't drawn any diagrams consisting of waves

 

[haruspex]

What do you know about reflected waves and standing waves?

 

[samisoccer9]

ive got the answer now , thanks tho

 

[berkeman]

Can you post your solution?

 

[samisoccer9]

Well it says EVERY 50CM, so I took that as the distance between nodes as its a standing wave question. I assumed the brick wall to be a fixed - open, so node to node, meaning LAMBDA is equal to 2L. L being 50cm --> (2)(.5) = 1 meter.
plug into universal wave equation v = f x lambda
343.2 = f x lambda
f = v/lamda --> 343.2/1
f = 343.2 Hz