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Whitepunk [10]
3 years ago
10

A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1

.80 m from the midpoint of the two speakers. What are the lowest two frequencies that produce an interference maximum at the microphone's location?
Physics
1 answer:
pantera1 [17]3 years ago
7 0

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x   = D + (y/2)

1st speaker distance from x   = 1.80 + (0.513/2) = 2.0565 m   .....1

and

at 2nd speaker distance from x   = D - (y/2)

2nd speaker distance from x   = 1.80 - (0.513/2) = 1.5435 m     .........2

so destructive interference from 1 and 2  we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = (  0+ 0.5) wavelength

wavelength  = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd  min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = (  1 + 0.5) wavelength

wavelength  = 0.342 m

frequency 2 =  velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

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In what way are gravitational and electrical forces similar?
Nadusha1986 [10]

Answer:

D. Both occur between objects independently whether they are in contact  or not.

Explanation:

- The gravitational force is a force that is exerted between two (or more) objects having mass. This force is always attractive and its magnitude is given by

F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 are the two masses, and r is the distance between the two masses.

- The electrical force is a force that is exerted between two (or more) objects having electrical charge. It can be either attractive or repulsive, depending on the sign of the two charges, and its magnitude is given by

F=k\frac{q_1 q_2}{r^2}

where k is the Coulomb's constant, q1 and q2 are the two charges, and r the distance between the two charges.

Looking at both formulas, we see that the two forces are present even when the two objects are not in contact with each other (in fact, r can assume any value in the formula). They are said to be non-contact forces. Therefore, the correct option is

D. Both occur between objects independently whether they are in contact  or not.

6 0
3 years ago
The amount of gravitational potential energy released as:_________.
nordsb [41]

Answer:

b.it depends on the distance it falls

8 0
3 years ago
A human hair is approximately 50 pm in diameter. Express this diameter in meters.
cupoosta [38]

A human hair is approximately 50 μm = 5 × 10⁻⁵ m  in diameter

4 0
3 years ago
Hope you all doing okay
Vsevolod [243]

Answer:Thank you

Explanation:

8 0
2 years ago
A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

5 0
3 years ago
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