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3 years ago

14

Find a model for simple harmonic motion that has an initial displacement (t=0) of 0 inches, an amplitude of 5 inches, and a freq

uency 4/3 cycles per second.

1 answer:

Svetach [21]3 years ago

7 0

Answer:

The equation of simple harmonic motion is $y=5\sin{(8\pi t)}$

Explanation:

Given that,

Amplitude = 5 inches

Frequency $f= \dfrac{4}{3}$

As per the question initial displacement is zero at t = 0 for sine curve.

The displacement is zero so the phase difference will be zero.

We know that,

The general equation of simple harmonic motion

$y=A\sin(\omega t+\phi)$

Where, A = amplitude

$\omega$ = angular frequency

$\phi$ = phase shift

We need to calculate the time period

Using formula of frequency

$f=\dfrac{1}{T}$

$T=\dfrac{1}{f}$

Put the value into the formula

$T=\dfrac{1}{\dfrac{4}{3}}$

$T=\dfrac{3}{4}$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

$\omega=\dfrac{2\pi\times4}{3}$

Put the value into the general equation

$y=5\sin{(8\pi t)}$

Hence, The equation of simple harmonic motion is $y=5\sin{(8\pi t)}$

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If the center of mass passes outside the area of support of an object, what will happen to it?

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Answer:

If a vertical line extending down from an object's CG extends outside its area of support, the object will topple

Explanation:

We can understand better this situation using a diagram with the forces acting on it.

In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.

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3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

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(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

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The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

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Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

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Extended by = 0.25mm = $0.25\times 10^{-3}$

Based on the above information, the calculation is as follows

a. The Stress of the wire is

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here area of circle = perpendicular to the are i.e cross-sectional i.e

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Now place these above values to the above formula

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= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

$= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}$

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