Two in-phase loudspeakers that emit sound with the same frequency are placed alo

Question

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.00 m. A person is standing 12.0 m away from the wall, equidistant from the loudspeakers. When the person moves 1.00 m parallel to the wall, she experiences destructive interference for the first time. What is the frequency of the sound? The speed of sound in air is 343 m/s. A) 211 Hz B) 256 Hz C) 422 Hz D) 512 Hz E) 614 Hz A 25.0-g suing is stretched with a tension of 43.0 N between two fixed points 12.0 m apart What is the frequency of the second harmonic? A) 6.00 Hz B) 12.0 Hz C) 18.0 Hz D)24.0 Hz E) 36 0 Hz

Explanation / Answer

10. let the frequency of the sound be f

speed of sound in air, v = 343 m/s

then wavelength of the sound, lambda = v/f

now, path diffrerence at the new location is t = sqroot(12^2 + 3.5^2) - sqroot(12^2 + 1.5^2) [ from the given data ]

so t = 0.4066133 m

so for first destructive interference, lambda/2 = t

so, v/f = 2*0.4066133

f = 421.7765 Hz

so option C)

11. MASS OF STRING, m = 0.025 kg

tension, T = 43 N

length, l = 12 m

frequency of nth harmonic, fn = v/lambda

where v is speed oif the wave on string and lambda is the wavelength

now, l = lambda [ for secoind harmonic]

and v = sqroot(Tl/m)

so, f = sqroot(T/ml) = sqroot(43/0.025*12) = 11.97 Hz

so option B) 12 Hz

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