A 0.181 kg mass is attached to a spring and executes simple harmonic motion with a pe-riod of 0.21 s. The total energy of the system is 1.2 J.

Find the force constant of the spring.

b) Find the amplitude of the motion.

total energy

1.2J
Well, when amplitude is max, it is all PEnergy, no KE.

f(t)=Asin(2PI/.21 * t)
v(t)=-A2PI/.21 * cos (2PIt/.21)
but when cos is =0, velocity is zero, and all PE is in spring
.21=1/2 kx^2 where x= A
solve for k

bobpursley, you just subbed in time for energy. I actually have the same question as JACK. and what you're saying doesn't make much sense to me. Also, we can't solve for k using that equation without x or A which we are looking for in part b.

To find the force constant of the spring, we can use the equation for the period of a mass-spring system:

T = 2π√(m/k)

Where T is the period, m is the mass, and k is the force constant of the spring.

In this case, T = 0.21 s and m = 0.181 kg. Rearranging the equation, we can solve for k:

k = (4π²m) / T²

Substituting the known values, we get:

k = (4π² * 0.181 kg) / (0.21 s)² ≈ 60.29 N/m

So the force constant of the spring is approximately 60.29 N/m.

To find the amplitude of the motion, we can use the equation for the total energy of a mass-spring system:

Total energy (E) = 1/2kA²

Where E is the total energy and A is the amplitude of motion.

In this case, E = 1.2 J and k = 60.29 N/m. Rearranging the equation, we can solve for A:

A = √(2E/k)

Substituting the known values, we get:

A = √(2 * 1.2 J / 60.29 N/m) ≈ 0.14 m

So the amplitude of the motion is approximately 0.14 m.