A block with mass m = 5.2 kg is attached to two springs with spring constants kl

Question

A block with mass m = 5.2 kg is attached to two springs with spring constants kleft = 32 N/m and kright = 48 N/m. The block is pulled a distance x = 0.26 m to the left of its equilibrium position and released from rest.
1.Where is the block located, relative to equilibrium, at a time 1.09 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value)
2. What is the net force on the block at this time 1.09 s? (a negative force is to the left; a positive force is to the right)
3. What is the total energy stored in the system?
4. If the block had been given an initial push, how would the period of oscillation change?

Explanation / Answer

1: To find where the block is located, we use the formula:

x(t) = -Acos(2t/T) where A is the distance and as per the given data distance=x=0.26m=A, t=time=1.09 s and T is the Time period , where T = 2(m/k) is euation (1)

keq = F / x and we calculate F= (kleft+kright)*Distance= (32+48)*0.26= 21 N

Keq= 21/0.26=81 N/m

Hence from Equation (1) T= 2*3.14 * (5.2/81)

T= 1.6 S

Hence as per x(t) = -Acos(2t/T)

x(t) = -0.26cos(2*3.14*1.09/1.6) = 0.108 m , which means block is in the right of equilibrium.

This completes the solution for 1.

Soulution for 2: F = -kx =-(32+48)*0.26= -20.8 N towards left

Solution 3:

Potential Energy becomes Kinetic Energy, or ½kx² = ½mv² =0.5(32+48)* 0.26^2= 2.704 J

Solution 4: Period would not change as it depends only on the equivalent spring constant and the mass.

Hope this helps you and good luck

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