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dedylja [7]
3 years ago
5

A child on a playground swings through a total of 32 degrees. If the displacement is equal on each side of the equilibrium posit

ion, what is the amplitude of this vibration?
Physics
1 answer:
Black_prince [1.1K]3 years ago
4 0

Answer: A = y/cos32

That is the amplitude A in terms of the displacement y.

Explanation: Since the displacement in the question In the question is the same in both direction, it is a Simple Harmonic Motion problem. In s.h.m the amplitude of displacement A is related to the displacement itself y by this simple equation

y = A* cos(theta)

So, A = y/cos(theta)

A = y/cos32.

If the magnitude of the displacement y is given, you just substitute in.

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(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa
Vedmedyk [2.9K]

Missing data: the wave number

k=60 cm^{-1}

(a)  z = 0.003 sin (6000y-31.4 t)

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

z = A sin (ky-\omega t)

where

A is the amplitude of the wave

k is the wave number

\omega is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

k = 60 cm^{-1} = 6000 m^{-1} is the wave number

T = 0.20 s is the period, so the angular frequency is

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s

So, the wave equation (in meters) is

z = 0.003 sin (6000y-31.4 t)

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

v_t = z' = -A \omega cos (ky-\omega t)

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

v_{t}_{max} =\omega A

where

\omega = 31.4 rad/s\\A = 0.003 m

Substituting,

v_{t}_{max}=(31.4)(0.003)=0.094 m/s

(c) 5.24 mm/s

The wave speed is given by

v=f \lambda

where

f is the frequency of the wave

\lambda is the wavelength

The frequency can be found from the angular frequency:

f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz

While the wavelength can be found from the wave number:

\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m

Therefore, the wave speed is

v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s

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3 years ago
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Answer:

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Explanation:

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The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the
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Answer: C. -1.16 meters/second2

Explanation:

A= v/t (velocity/time)

in this case: v=7 and t=6

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Answer:

5.714 hours / day

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