A system consists of a block of mass m attached to a spring of negligible mass a
ID: 1641569 • Letter: A
Question
A system consists of a block of mass m attached to a spring of negligible mass and force constant k. The block is free to move on a frictionless horizontal surface, while the left end of the spring is held fixed. When the spring is neither compressed nor stretched, the block is in equilibrium. If the spring is stretched, the block is displaced to the right and when it is released, a force acts on it to pull it back toward equilibrium. By the time the block has returned to the equilibrium position, it has picked up some kinetic energy, so it overshoots, stopping somewhere on the other side, where it is again pulled back toward equilibrium. As a result, the block moves back and forth from one side of the equilibrium position to the other, undergoing oscillations. Since we are ignoring friction (a good approximation to many cases), the mechanical energy of the system is conserved and the oscillations repeat themselves over and over. a) 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). The initial acceleration of the block, a0, when the block is released at a distance A from its equilibrium position is equal to -kA/m. What is the acceleration, a, 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
The answer is zero.
When a body undergoes simple harmonic motion with initial displcement A from the equilibrium position, its motion is decribed by the equation
x(t)=Acos wt
A=initial displacement
w = angular velocity of motion
acceleration = d2x/dt2 = -Aw2coswt
When the mass is passing equilibrium position, x(t)=0, and hence cos wt = 0(becasue A is not zero)
Therefore, acceleration at that time=-Aw2 cos wt = 0
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