Question

Two loudspeakers in a plane, 5.0 m apart, are playing the same frequency. If you stand 12.0 m in front of the plane of the speak- ers, centered between them, you hear a sound of maximum in- tensity. As you walk parallel to the plane of the speakers, staying 12.0 m in front of them, you first hear a minimum of sound inten- sity when you are directly in front of one of the speakers. What is the frequency of the sound? Assume a sound speed of 340 m/s

Answer 1

Answer:

The frequency of the sound is 170 Hz

Explanation:

Traveling wave is given as;

$D(r,t) = D_o(r)sin(kr- \omega t + \phi_o)$

where;

r is the distance from the source

$(kr- \omega t + \phi_o)$ is phase

Then phase difference is given as;

$\delta \phi = 2 \pi\frac{\delta r}{\lambda} +\delta \phi_o$

The phase difference between the two speakers at maximum intensity;

$\delta \phi = 2\pi \frac{\delta r}{\lambda}$

where;

λ is the wavelength

Δr is difference in distance between the two speakers

Δr = r₂ - r₁ = $\sqrt{L^2 + d^2} -L$

Given;

distance between the two speakers, d = 5.0 m

distance to the plane of the speakers, L = 12.0 m

Δr = $\sqrt{12^2 + 5^2} -12 = 1 \ m$

$\delta \phi = 2\pi\frac{\delta r}{\lambda}$

$\delta r = \frac{\delta \phi}{2\pi}\lambda$, at minimum sound intensity, ΔФ = π

$\delta r = \frac{\pi}{2\pi} \lambda\\\\\delta r = \frac{\lambda}{2}$

λ = 2Δr

λ = 2 (1) = 2m

$f = \frac{v}{\lambda} = \frac{340}{2} = 170\ Hz$

Therefore, the frequency of the sound is 170 Hz

Answer 2

Answer:

170 Hz

Explanation:

The path difference between two waves are given by:

Δ=$\sqrt{12^2 +5^2 }$ - 12

Δ = 1 m

In order to hear minimum intensity, the path difference will be

Δ = λ/2

λ = 2Δ

Also, We know that

V= f λ

λ=V/f

Therefore,

340/f = 2 x 1

f= 170 Hz

Thus, the frequency of the sound is 170 Hz