Q)Two loudspeakers, A and B, are driven by the same amplifier as shown and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. a) What is the lowest frequency for which constructive interference occurs at point Q? b) What is the lowest frequency for which destructive interference occurs at point Q? The speed of sound wave is 344 m/s.
Answer:
lowest frequency is 172 Hz. for n = 1
lowest frequency is 86 Hz for n = 0
Explanation:
(a) Constructive interference takes place when the path difference is n where n=1,2,3 ...
frequency f = v /λn
= v n / d
= n ( 344 / 2 )
= n ( 172 ) Hz.
lowest frequency is 172 Hz
b)Destructive interference takes place when the path difference is n/2 x where n=1,3,5 ...
λ = 4/2n+1
for n = 0
λ = 4m
f = 344/4
f = 86 Hz.
lowest frequency is 86 Hz
Answer:
320 N/m
Explanation:
From Hooke's law, we deduce that
F=kx where F is applied force, k is spring constant and x is extension or compression of spring
Making k the subject of formula then
Conversion
1m equals to 100cm
Xm equals 25 cm
25/100=0.25 m
Substituting 80 N for F and 0.25m for x then
Therefore, the spring constant is equal to 320 N/m
Answer:
619.8 N
Explanation:
The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:
where
T is the tension
m is the mass of the rock
v is the speed
r is the radius of the circular path
At the beginning,
T = 50.4 N
v = 21.1 m/s
r = 2.51 m
So we can use the equation to find the mass of the rock:
Later, the radius of the string is decreased to
r' = 1.22 m
While the speed is increased to
v' = 51.6 m/s
Substituting these new data into the equation, we find the tension at which the string breaks:
Answer:
It changes at a rate of 4/3 meter per second
Explanation:
In the given figure below we have
Solving for Y given we get
