Question

An oscillating bock-spring system has a mechanical energy of 1.0 J, an amplitude of 0.10 m, and a maximum speed of 1.2 m/s. Find (a) the force constant of the spring, (b) the mass, and (c) the frequency of oscillation.

Answer 1

a)
`k = 200.016\,(N)/(m)`, b)
`m = 1.389\,kg`, c)
`f = 0.524\,hz`

Step-by-step explanation:

a) The maximum speed of the oscillating block-spring system is:

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

The angular frequency is:

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

`\omega = (1.2\,(m)/(s) )/(0.1\,m)`

`\omega = 12\,(rad)/(s)`

The mass of the system is:

`E = (1)/(2)\cdot m\cdot v_(max)^(2)`

`m = (2\cdot E)/(v_(max)^(2))`

`m = (2\cdot (1\,J))/((1.2\,(m)/(s) )^(2))`

`m = 1.389\,kg`

The spring constant is:

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

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

`k = (12\,(rad)/(s) )^(2)\cdot (1.389\,kg)`

`k = 200.016\,(N)/(m)`

b) The mass is:

`m = 1.389\,kg`

c) The frequency of oscillation is:

`\omega = 2\pi\cdot f`

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

`f = (2\pi)/(12\,(rad)/(s) )`

`f = 0.524\,hz`

Answer 2

a) F = 20 N

b) m = 1.39 kg

c) f = 1.909 Hz

Step-by-step explanation:

Given

E = 1 J

A = 0.1 m

vmax = 1.2 m/s

a) F = ?

b) m = ?

c) f = ?

Solution

a) We apply the equation

E = 0.5*k*A²

then

k = 2*E/A²

k = 2*1 J/(0.1 m)²

k = 200 N/m

then we use the equation

F = kA

F = (200 N/m)(0.1 m)

F = 20 N

b) We use the formula

E = K + U

if U = 0 J

then

E = K = 0.5*m*v²

⇒ m = 2*K/v²

m = 2*1 J/(1.2 m/s)²

m = 1.39 kg

c) we apply the equation

f = (1/2π)√(k/m)

then

f = (1/2π)√(200 N/m/1.39 kg)

f = 1.909 Hz