Question

A 0.36 kg object, attached to a spring with constant k=10n/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude 0.082 m. What is it speed when it’s displacement is 0.041 m

Answer 1

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

Step-by-step explanation:

First, we need to determine the angular frequency of the system (
`\omega`), in radians per second:

`\omega = \sqrt{(k)/(m) }` (1)

Where:

`k` - Spring constant, in newtons per meter.

`m` - Mass, in kilograms.

If we know that
`k = 10\,(N)/(m)` and
`m = 0.36\,kg`, then the angular frequency of the system is:

`\omega = \sqrt{(10\,(N)/(m) )/(0.36\,kg) }`

`\omega \approx 5.270\,(rad)/(s)`

The kinematic formulas for the position (
`x(t)`), in meters, velocity (
`\dot x(t)`), in meters per second, and acceleration of the object (
`\ddot x(t)`), in meters per square second, are:

`x(t) = A\cdot \cos \omega t` (2)

`\dot x(t) = -\omega \cdot A \cdot \sin \omega t` (3)

`\ddot x(t) = -\omega^(2)\cdot A \cdot \cos \omega t` (4)

Where
`A` is the amplitude of the motion, in meters.

From (2) we determine the time associated with position
`x(t) = 0.041\,m` (
`\omega \approx 5.270\,(rad)/(s)`,
`A = 0.082\,m`):

`t = (1)/(\omega)\cdot \cos^(-1) \left((x(t))/(A) \right)` (5)

`t = (1)/(5.270\,(rad)/(s) )\cdot \cos^(-1)\left((0.041\,m)/(0.082\,m) \right)`

`t = 0.199\,s`

And the speed of the object is:

`\dot x(t) = -\left(5.270\,(rad)/(s) \right)\cdot (0.082\,m)\cdot \sin \left[\left(5.270\,(rad)/(s) \right)\cdot (0.199\,s)\right]`

`\dot x(t) \approx -0.375\,(m)/(s)`

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.