A 2.0-kg mass attached to a spring is displaced 8.0 cm from equilibrium position.It is released and it oscillates with a frequency of 4.0 Hz.What is the speed of the block when it is 6.5 cm from equilibrium? Assume that the surface is frictionless.
A)1.08 m/s
B)1.17 m/s
C)2.11 m/s
D)3.44 m/s
E)3.72 m/s

Question

Quizplus · Accepted Answer

First, we need to find the amplitude, A, of the oscillation. We know that the mass is displaced 8.0 cm from equilibrium, so A = 8.0 cm = 0.08 m.

Next, we can use the formula for the velocity of an oscillating mass: 
v = ±ω√(A^2 - x^2)

where ω is the angular frequency (2πf), and x is the displacement at any given time. To find the velocity when the mass is 6.5 cm from equilibrium, we plug in:

v = ±2π(4.0 Hz) √(0.08^2 - 0.065^2)
v = ±1.17 m/s

Note that we use the negative sign in front of the square root because the mass is moving towards equilibrium, so its velocity should be negative. Therefore, the answer is B, 1.17 m/s.

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