Question

A 92 g particle undergoes SHM with an amplitude of 1.7 mm, a maximum acceleration of magnitude 7.0 x 103 m/s2, and an unknown phase constant φ. What are

(a) the period of the motion,
(b) the maximum speed of the particle, and
(c) the total mechanical energy of the oscillator?

What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?

Answer 1

a)
`T = 3.097\,s`, b)
`v_(max) = 3.449* 10^(3)\,(m)/(s)`, c)
`E = 5.477* 10^(-7)\,J`, d)
`F = 6.443* 10^(-4)\,N`, e)
`F = 3.222* 10^(-4)\,N`

Step-by-step explanation:

a) The angular frequency is:

`\omega = \sqrt{(a_(max))/(A)}`

`\omega = \sqrt{(7* 10^(3)\,(m)/(s^(2)) )/(1.7* 10^(-3)\,m) }`

`\omega = 2.029\,(rad)/(s)`

The period of the motion is:

`T = (2\pi)/(\omega)`

`T = (2\pi)/(2.029\,(rad)/(s) )`

`T = 3.097\,s`

b) The maximum speed of the particle is:

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

`v_(max) = \left(2.029\,(rad)/(s)\right)\cdot (1.7\cdot 10^(-3)\,m)`

`v_(max) = 3.449* 10^(3)\,(m)/(s)`

c) The spring constant is:

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

`k = (2.029\,(rad)/(s) )^(2)\cdot (0.092\,kg)`

`k = 0.379\,(N)/(m)`

The total mechanical energy of the oscillator is:

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

`E = (1)/(2)\cdot \left(0.379\,(N)/(m)\right)\cdot (1.7* 10^(-3)\,m)^(2)`

`E = 5.477* 10^(-7)\,J`

d) The magnitude of the force on the particle at its maximum displacement is:

`F = k\cdot A`

`F = (0.379\,(N)/(m))\cdot (1.7* 10^(-3)\,m)`

`F = 6.443* 10^(-4)\,N`

e) The magnitude of the force on the particle at half of its maximum displacement is:

`F = (1)/(2)\cdot k\cdot A`

`F = (0.379\,(N)/(m))\cdot (0.85* 10^(-3)\,m)`

`F = 3.222* 10^(-4)\,N`