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rjkz [21]
3 years ago
5

A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 205 Hz. A person on the pl

atform right next to the speaker detects the sound waves reflected off the wall and those emitted by the speaker. How fast should the platform move, vp , for the person to detect a beat frequency of 5.00 Hz?
Take the speed of sound to be 344 m/s.
Physics
1 answer:
Arlecino [84]3 years ago
6 0

Answer:

Explanation:

The question relates to Doppler effect and beat.

The observer is moving towards the reflected sound so apparent frequency will be increased

f = f₀ x (V + v₁) / (V - v₂)

f is apparent frequency , f₀ is real frequency , V is velocity of sound , v₁ is velocity of observer and v₂ is velocity of source . Here

v₁ = v₂ = vp as both observer and source have same velocity

f = f₀ x (V + v₁) / (V - v₂)

205 +5 = 205 x (344 +vp)/ ( 344 - vp)

1.0234 = (344 +vp)/ ( 344 - vp)

= 352 - 1.0234vp = 340+vp

12 / 2.0234vp

vp = 6 m /s approx.

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If we will connect the resistors 2ohms, 3ohms, 5ohms in series and the 10ohms resistance parallel then we get equivalent resistance of 5 ohms.

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Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

v^{2} = \frac{dr}{dt}^{2}  +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2} - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

\frac{r}{z}=tan(45)

or:

z = r cot(45) - (2)

and:

\frac{dz}{dt} = \frac{dr}{dt} cot(45)

replacing (2) in (1) we obtain:

v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}  - (3)

Now the kinetic energy is given as:

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Obtained from the Euler-Langrange equations

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mr\frac{d\theta}{dt}=c

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