In an experiment to demonstrate the interference odf sound waves, two loudspeakers s1 and s2 are connected to a signal generator.Ken stands in front of te loudspeaker and then walks from points X to pointY.
AS he walks , he hears the first and the second loud sound at point O and P respectively . Form point P to point Y , he hears a soft sound at point Q .it is given that s1O=s2O
圖片參考：http://imgcld.yimg.com/8/n/HA08381517/o/701205090043013873404801.jpg
1.)calculate the path diference at point Q
2.)what is the wavelength of the sound?
3.)what is the frequency of the sound?
p.s: take the spped of sound in air to be 340ms-1
phy help again plz
2012-05-10 12:38 am
回答 (2)
2012-05-10 5:30 pm
✔ 最佳答案
圖片參考：http://imgcld.yimg.com/8/n/HA08205231/o/701205090043013873404810.jpg
1. Because S1O = S2O,
S1S2= 1M,
So, AO = BO = 0.5M,
As we know, the distance between S2 & XY is S1A = S2B = 6M,
QO = 1.5M,
So, QA = QO-AO = 1M and QB = QO+BO = 2M,
The path diff. = S2Q - S1Q
= w(S2B^2+QB^2)-w(S1A^2+QA^2)
= 0.242 (M)
2. According to the interference theory of sound,
The wavelength = 0.242 x 2 / 3
= 0.161 (M)
3. The frequency of the sound =
sound speed/wavelength = 340/0.161 = 2110 (Hz)
2012-05-10 3:37 am
1. Let a be the angular position of point Q from the central axis of the two loudspeakers.
tan(a) = 1.5/6
i.e. a = 14.04 degrees
Hence, path difference = 1 x sin(14.04) m = 0.2425 m
2. Since a soft sound is heard at Q, destructive interference occurs there.
hence, 0.2425 = 1.5入
where 入 is the wavelength of sound
入 = 0.1616 m
3. Frequency of sound = 340/0.1616 Hz = 2100 Hz
tan(a) = 1.5/6
i.e. a = 14.04 degrees
Hence, path difference = 1 x sin(14.04) m = 0.2425 m
2. Since a soft sound is heard at Q, destructive interference occurs there.
hence, 0.2425 = 1.5入
where 入 is the wavelength of sound
入 = 0.1616 m
3. Frequency of sound = 340/0.1616 Hz = 2100 Hz
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