Question

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, and an unknown phase constant ?.(a) What is the period of the motion?1 s(b) What is the maximum speed of the particle?2 m/s(c) What is the total mechanical energy of the oscillator?3 J(d) What is the magnitude of the force on the particle when the particle is at its maximum displacement?4 N(e) What is this force when the particle is at half its maximum displacement?5 N

Answer 1

a)
`T=0.0031416s`

b)
`v_(max)=6.283(m)/(s)`

c)
`E=0.1974J`

d)
`F=80N`

e)
`F=40N`

Step-by-step explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by
`X=Acos(\omega t +\phi)` (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

`(dX)/(dt)=v(t)=-A\omega sin(\omega t +\phi)` (2)

For the acceleration

`(d^2 X)/(dt)=a(t)=-A\omega^2 cos(\omega t+\phi)` (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

`A\omega^2=8x10^(3)(m)/(s^2)`

Since we know the amplitude A=0.002m we can solve for
`\omega` like this:

`\omega =\sqrt{(8000(m)/(s^2))/(0.002m)}=2000(rad)/(s)`

And we with this value we can find the period with the following formula

`T=(2\pi)/(\omega)=(2 \pi)/(2000(rad)/(s))=0.0031416s`

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

`v_(max)=A\omega =0.002mx2000(rad)/(s)=4(m rad)/(s)=4(m)/(s)`

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

`E=(1)/(2)mv^2_(max)=(1)/(2)(0.01kg)(6.283(m)/(s))^2=0.1974J`

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens
`cos(\omega t +\phi)=1`, and we also know that the maximum acceleration is given by::

`|(d^2X)/(dt^2)|=A\omega^2`

So then we have that:

`F=ma=mA\omega^2`

And since we have everything we can find the force

`F=ma=0.01Kg(0.002m)(2000(rad)/(s))^2 =80N`

6) Part e

When the mass it's at the half of it's maximum displacement the term
`cos(\omega t +\phi)=1/2` and on this case the acceleration would be given by;

`|(d^2X)/(dt^2)|=A\omega^2 cos(\omega t +\phi)=A\omega^2 (1)/(2)`

And the force would be given by:

`F=ma=(1)/(2)mA\omega^2`

And replacing we have:

`F=(1)/(2)(0.01Kg)(0.002m)(2000(rad)/(s))^2 =40N`