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Gala2k [10]
3 years ago
12

A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , wher

e x is in cm and t is in seconds. What is the frequency of the oscillation?
Physics
2 answers:
zhuklara [117]3 years ago
7 0

Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

x(t) = Acos(\omega t - \phi)

where A is the amplitude, \omega = 2\pi f is the angular frequency and \phi is the initial phase.

Since our body has an equation of  x = 5cos(π t + π/3) we can equate \omega = \pi and solve for frequency f

2\pi f = \pi

f = 1/2 Hz

IRISSAK [1]3 years ago
6 0
<h2>Answer:</h2>

0.5Hz

<h2>Explanation:</h2>

The general equation of the displacement, x, of a body undergoing simple harmonic motion at a given point in time (t) is given by;

x = A cos (ωt ± ∅)  --------------------------(i)

where;

A = amplitude of the wave

ω = angular velocity of the wave

∅ = phase constant of the wave

<em>From the question;</em>

x = 5cos(π t + π/3)      -----------------------------(ii)

<em>Comparing equations (i) and (ii), the following deductions among others can be made;</em>

A = 5cm

ω = π

<em>But the angular velocity (ω) of the wave is related to its frequency (f) as follows;</em>

ω = 2 π f        --------------------(iii)

<em>Substitute the value of ω = π  into equation (iii) as follows;</em>

π = 2 π f

<em>Divide through by π;</em>

1 = 2f

<em>Solve for f;</em>

f = 1/2

f = 0.5

Frequency (f) is measured in Hz. Therefore, the frequency of the oscillation is 0.5Hz

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Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz
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(C) The frequency decrease and intensity decrease

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The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

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The frequency is the inverse from the wavelength, so the frequency heard will increase.

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A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?​
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<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg

k = 0.4

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F= kMA

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2 years ago
A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
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(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

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