A block with a mass of 0.800 kg is connected to a spring, displaced in the posit

Question

                    A block with a mass of 0.800 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from the equilibrium, and released from                     rest at t=0. The block then oscillates without friction on a horizontal surface. After being released, the first time the block is a distance of 35.0cm from                     the equilibrium is at t=0.200s.
                

                    (a) what is the block's period of oscillation?                 

                    (b)What is the value of the spring constant?                 

                    (c) what is the block's velocity at t=0.200 s? (indicate direction with sign)                 

                    (d) what is the block's acceleration at t=0.200 s? (indicate direction with sign)                 

Explanation / Answer

y = ASin(wt)

As we know

w = sqrt(k/m)

Therefore

y = ASin(sqrt(k/m)*t)

0.35 = 0.5*Sin(sqrt(k/0.8) * 0.2)

k = 39475.17 N/m

T = 2pi*sqrt(m/k)

= 2pi*sqrt(0.800/39475.17)

= 0.0282 sec

Here Total Energy remains Constant

Therefore

0.5*k*A^2 = 0.5*m*v^2 + 0.5*k*x^2

0.5*39475.17*0.5^2 = 0.5*39475.17*0.35^2 + 0.5*0.8*v^2

v = 79.318 m/sec

Acceleration = - w^2A*Cos(wt)

= - (0.800/39475.17)*0.5*Cos((0.800/39475.17)*0.200)

= - 1.01*10^-5 m/sec^2

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