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Olenka [21]
2 years ago
10

The displacement x with respect to time t of a particle moving in simple harmonic motion is given by x = 5cos(16pi*t). where x i

s in mm and t is in seconds. If the particle starts at x = 5mm and t = 0s. At what time t does it first pass through its equilibrium position?
A) 1/32 s
B) 1/16 s
C) 1/5 s
D) 4 s
E) 8 s
Physics
2 answers:
Kitty [74]2 years ago
5 0

Answer:

A that's is the answer

letter A is the answer

Strike441 [17]2 years ago
4 0
T
x=0.05sin6t
(a) the amplitude of the oscillations
A
=
0.05
m
A=0.05m
the period of oscillations
T
=
2
π
ω
=
2
π
6
=
2.1
s
T=
ω
2π
​
=
6
2π
​
=2.1s
the maximum acceleration
a
max
⁡
=
A
ω
2
=
0.05
∗
6
2
=
1.8
m
/
s
2
a
max
​
=Aω
2
=0.05∗6
2
=1.8m/s
2

(b)
x
¨
+
ω
2
x
=
0
x
¨
+ω
2
x=0
ω
=
k
/
m
ω=
k/m
​
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b ) spectral width in terms of wavelength = 30 nm

frequency width = ?

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c )

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50 dB = 5 B

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I = I₀ x 10⁵

= 10⁻¹² x 10⁵

= 10⁻⁷ W

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5 0
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Yuliya22 [10]

Answer:

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Since their frequencies are similar, we should have beats of high and low frequency.

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Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.

So, as you wander around the room, I should hear points of high and low sound across the room.

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