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1 year ago

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A transverse sinusoidal wave on a string has a period T= 25.0ms and travels in the negative x direction with a speed of 30.0 m/s

. At t = 0 , an element of the string at x = 0 has a transverse position of 2.00cm and is traveling downward with a speed of 2.0 m/s .(a) What is the amplitude of the wave?

1 answer:

svlad2 [7]1 year ago

5 0

The amplitude of the wave is 0.022m

<h3 /><h3>How to solve solve for the amplitude</h3>

<u>Given data</u>

period T= 25.0ms = 0.025s

speed of 30.0 m/s

t = 0

x = 0

transverse position of 2.00cm = 0.02m

<h3 />

wave equation is = y ( x ) = A cos ( ωx + ∅ )

since:

t = 0

x = 0

0.02 = A cos ( 0 + ∅ )

0.02 = A cos ( ∅ )

speed = v ( x )= 2.0 m/s = d y ( x ) / d x

- 2 = - Aω sin ( ∅ )

dividing the equations as below gives

- Aω sin ( ∅ ) / A cos ( ∅ )

- ω tan ( ∅ ) = - 2 / 0.02

ω tan ( ∅ ) = 100

ω = 2 * pi * f = 2 * pi / T = 2 * pi / 0.025 = 80 * pi

tan ( ∅ ) = 100 /( 80* pi )

tan ( ∅ ) = 0.398

∅ = 0.3787 rad

Amplitude = 0.3787 / cos ∅

Amplitude = 0.022m

Read more on amplitude of wave here: brainly.com/question/19036728

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We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

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Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

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I hope it helps you!

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