As shown in the figure(Figure 2) , a coordinate system with the origin at the eq

Question

As shown in the figure(Figure 2) , a coordinate system with the origin at the equilibrium position is chosen so that the x coordinate represents the displacement from the equilibrium position. (The positive direction is to the right.) What is the initial acceleration of the block, a0, when the block is released at a distance A from its equilibrium position?

What is the acceleration a1 of the block when it passes through its equilibrium position?

Express your answer in terms of some or all of the variables A, m, and k.

Explanation / Answer

here,

Spring Force, Fs
spring Constant, k

Spring force = Spring Constant * Extension
fs = k *x

max dispalcement, x = A
At equillibrium displacement will be ,x = 0

Part A:
From newtom Law of motion , Fs = Fnet
Fs = mass * acceleration
k*A = m*ao

ao = K*A/m

Part B:

acceleration at Equillibrium will be zero as x = 0

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.