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2 years ago

What happens to the amplitude of the resultant wave when two sound waves with equal amplitude constructively interfere?

2 answers:

7 0

Constructive interference means that the waves combine instead of negate. if they are both amplitude x, adding them together would result in 2x, or doubling the amplitude of one of the waves

Alinara [238K]2 years ago

4 0

Answer:

The amplitude of the resultant wave will be twice the amplitude of each wave

Explanation:

- Constructive interference occurs when two waves meet in phase ( $\theta=0^{\circ}$ , which means that the crest of a wave meets the crest of the other wave: in this case, the amplitude of the resultant wave is the sum of the amplitudes of the single waves

- Destructive interference occurs when the two waves meet in anti-phase ( $\theta=180^{\circ}$ , which means that the crest of a wave meets the trough of the other wave: in this case, the amplitude of the resultant wave is the difference between the amplitudes of the single wave

In this problem, we have constructive interference between the two waves: this means that the resultant amplitude is the sum of the two waves. Since the two waves have equal amplitude (we can call it A), the resultant amplitude will be

$R= A+A=2 A$

so, twice the amplitude of the single wave.

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1 year ago

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2 years ago

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3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago