Question

A 0.227-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.279 m and x2 = 0.469 m. The period of oscillation is 0.609 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.

Answer 1

a)
`f = 1.642\,hz`, b)
`\bar x = 0.095\,m`, c)
`A = 0.374\,m`, d)
`v_(max) = 3.859\,(m)/(s)`, e)
`a_(max) = 39.809\,(m)/(s^(2))`, f)
`k = 24.162\,(N)/(m)`, g)
`E = 1.690\,J`

Step-by-step explanation:

a) The frequency is the reciprocal of the period of oscillation:

`f = (1)/(T)`

`f = (1)/(0.609\,s)`

`f = 1.642\,hz`

b) The equilibrium position is the average of the extreme points:

`\bar x = (-0.279\,m +0.469\,m)/(2)`

`\bar x = 0.095\,m`

c) The amplitude is the absolute of the substraction of the equilibrium position from any of the extreme points:

`A = |0.469\,m - 0.095\,m|`

`A = 0.374\,m`

d) The angular frequency is:

`\omega = 2\pi \cdot f`

`\omega = 2\pi \cdot (1.642\,hz)`

`\omega \approx 10.317\,(rad)/(s)`

The maximum speed is:

`v_(max) = \omega \cdot A`

`v_(max) = (10.317\,(rad)/(s) )\cdot (0.374\,m)`

`v_(max) = 3.859\,(m)/(s)`

e) The maximum acceleration is:

`a_(max) = \omega^(2)\cdot A`

`a_(max) = (10.317\,(rad)/(s) )^(2)\cdot (0.374\,m)`

`a_(max) = 39.809\,(m)/(s^(2))`

f) The force constant is:

`k = \omega^(2)\cdot m`

`k = (10.317\,(rad)/(s) )^(2)\cdot (0.227\,kg)`

`k = 24.162\,(N)/(m)`

g) The total mechanical energy is:

`E = (1)/(2)\cdot k \cdot A^(2)`

`E = (1)/(2)\cdot (24.162\,(N)/(m) )\cdot (0.374\,m)^(2)`

`E = 1.690\,J`