Question

A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985 m and a duration of 127 s for 76 cycles of oscillation.

Find the frequency, f, the speed at the equilibrium position, vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 38.1% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position.

Answer 1

a)
`f = 0.598\,hz`, b)
`v_(max) = 3.701\,(m)/(s)`, c)
`k = 15.385\,(N)/(m)`, d)
`U = 1.081\,J`, e)
`K = 6.382\,J`, f)
`v\approx 3.422\,(m)/(s)`

Step-by-step explanation:

a) The frequency of oscillation is:

`f = (76)/(127\,hz)`

`f = 0.598\,hz`

b) The angular frequency is:

`\omega = 2\pi \cdot f`

`\omega = 2\pi \cdot (0.598\,hz)`

`\omega = 3.757\,(rad)/(s)`

Lastly, the speed at the equilibrium position is:

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

`v_(max) = (3.757\,(rad)/(s) )\cdot (0.985\,m)`

`v_(max) = 3.701\,(m)/(s)`

c) The spring constant is:

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

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

`k = (3.757\,(rad)/(s) )^(2)\cdot (1.09\,kg)`

`k = 15.385\,(N)/(m)`

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

`U = (1)/(2)\cdot (15.385\,(N)/(m) )\cdot (0.375\,m)^(2)`

`U = 1.081\,J`

e) The maximum potential energy is:

`U_(max) = (1)/(2)\cdot (15.385\,(N)/(m) )\cdot (0.985\,m)^(2)`

`U_(max) = 7.463\,J`

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

`K = U_(max) - U`

`K = 7.463\,J - 1.081\,J`

`K = 6.382\,J`

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

`K = (1)/(2)\cdot m \cdot v^(2)`

`v = \sqrt{(2\cdot K)/(m) }`

`v = \sqrt{(2\cdot (6.382\,J))/(1.09\,kg) }`

`v\approx 3.422\,(m)/(s)`