A .530 kg block slides on a frictionless horizontal surface with a speed of 1.10 m/s. The block encounters an unstretched spring
1 answer:
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Answer:
a) 343.0 Hz b) 686.0 Hz
Explanation:
a) First, we need to know the distance to both speakers.
If the person is at halfway between the two speakers, and they are 4.0 m apart, this means that he is at 2.0 m from each speaker.
So, if he moves 0.25 m towards one of them, the distance from any speaker will be as follows:
d₁ = 2.0 m-0.25 m= 1.75 m
d₂ = 2.0 m + 0.25 m = 2.25 m
The difference between these distances is the path difference between the sound from both speakers:
d = d₂ - d₁ = 2.25 m - 1.75 m = 0.5 m
If the person encounters at this path difference a minimum of sound intensity, this means that this distance must be an odd multiple of the semi-wavelength:
d = (2*n-1)*(λ/2) = 0.5 m
The minimum distance is for n=1:
⇒ λ = 2* 0.5 m = 1 m
In any wave, there exists a fixed relationship between the speed (in this case the speed of sound), the wavelength and the frequency, as follows:
v = λ*f, where v= 343 m/s and λ=1 m.
Solving for f, we have:
b) If the person remains at the same point, for this point be a maximum of sound intensity, now the path difference (that it has not changed) must be equal to an even multiple of the semi-wavelength, which means that it must be met the following condition:
d = 0.5 m = 2n*(λ/2) = λ (for n=1)
if the speed remains the same (343 m/s) we can find the new frequency as follows:
⇒ f = 686.0 Hz