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A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. . Assume that the +xdirection is to the right. The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero initial velocity. Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion.

 the system will undergo simple harmonic motion. For such a system, the equation of motion is a(t)=-\frac{k}{m}x(t),

and its solution, which provides the equation for x(t), is  x(t)=A\cos\left(\sqrt{\frac{k}{m}}t\right).

 

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ParA: At what time t_1 does the block come back to its original equilibrium position (x=0) for the first time?
Express your answer in terms of some or all of the variables: Ak, and m.
 

Part B
Find the velocity v of the block as a function of time.
Express your answer in terms of some or all of the variables: kmA, and t.
 

Part C
What is the maximum speed v_max of the block during this motion?
Express your answer in terms of some of all of the variables: km, and A.

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B. Haranadh Kumar
B. Haranadh KumarLv10
7 Nov 2020
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