I attach a 4.0-kg block to a spring that obeys Hooke's law and supply 3.4 J of energy to stretch the spring.I release the block;it oscillates with period 0.15 s.The amplitude is
A)2.2 cm.
B)3.1 cm.
C)1.6 cm.
D)20 cm.
E)4.4 cm.

Question

Quizplus · Accepted Answer

We can start by using the formula for the period of a mass-spring system:
T = 2π √(m/k)
where T is the period, m is the mass attached to the spring, and k is the spring constant.

From this formula, we can solve for the spring constant:
k = (2π/T)^2  * m

k = (2π/0.15)^2 * 4.0 = 688.7 N/m

Now we can use the energy equation:
E = (1/2) k x^2

where E is the energy supplied to the spring and x is the amplitude of the oscillation.

Rearranging this equation to solve for x, we get:
x = √(2E/k)

x = √(2 * 3.4 / 688.7) = 0.0311 m = 3.1 cm

Therefore, the amplitude of the oscillation is 3.1 cm, so the correct answer is B.

I Attach a 4