Question

If the displacement, velocity and acceleration at an instant of a particledescribing S.H.M are respectively 7.5m, 7.5m/s, 7.5m/s?. Calculate themaximum velocity of the particle.

Answer 1

We are given the displacement, velocity, and acceleration of a particle that describes Simple Harmonic Motion. We are asked to determine the maximum velocity of the particle. To do that we can use the following equation for the maximum velocity of a particle describing SHM:

`v_(máx)=-A\omega`

Where:

`\begin{gathered} A=\text{ amplitude} \\ \omega=\text{ angular frequency} \end{gathered}`

To determine the values of amplitude and angular frequency we can use the expressions for displacement, velocity, and acceleration. The expression for displacement is:

`x=A\cos (\omega t)`

The expression for the velocity is:

`v=-A\omega\sin (\omega t)`

And the expression for acceleration is:

`a=-A\omega^2\cos (\omega t)`

Now, from the expression for the displacement we can solve for the amplitude, like this:

`(x)/(\cos(\omega t))=A`

Now we can replace this in the expression got eh acceleration:

`a=-(x)/(\cos(\omega t))\omega^2\cos (\omega t)`

Simplifying:

`a=-x\omega^2`

Now we can solve for the angular frequency:

`-(a)/(x)=\omega^2`

Taking square root to both sides:

`\sqrt[]{-(a)/(x)}=\omega`

replacing the values:

`\sqrt[]{-(-7.5(m)/(s^2))/(7.5m)}=\omega`

Solving the operations:

`1s^(-1)=\omega`

Now we divide the formula for the displacement and the formula for the velocity, we get:

`(v)/(x)=(-A\omega\sin (\omega t))/(A\cos (\omega t))`

Simplifying we get:

`(v)/(x)=-\omega\tan (\omega t)`

Replacing the known values:

`(-7.5(m)/(s))/(7.5m)=-(1s^(-1))\tan (t)`

Simplifying:

`-1=-\tan (t)`

Solving for "t":

`t=\arctan (1)=0.78`

Now we can replace these values in the formula for displacement to get the value of the amplitude:

`x=A\cos (\omega t)`

Solving for the amplitude:

`(x)/(\cos (\omega t))=A`

Replacing the known values:

`(7.5m)/(\cos (0.78))=A`

Solving the operations:

`10.6m=A`

Now we replace the values in the formula for the maximum velocity:

`v_{\text{max}}=A\omega`

Replacing:

`\begin{gathered} v_(\max )=(10.6m)(1s^(-1)) \\ v_(\max )=10.6\text{ m/s} \end{gathered}`

Therefore, the maximum velocity is 10.6 meters per second.