A 1.6-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m.The block is pulled from its equilibrium position at to a displacement x = +0.080 m and is released from rest.The block then executes simple harmonic motion along the horizontal x-axis.What is the velocity of the block at time ?
A) 0.82 m/s
B) -0.82 m/s
C) 0.30 m/s
D) -0.30 m/s
E) 0.00 m/s

Question

A 1.6-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m.The block is pulled from its equilibrium position at  to a displacement x = +0.080 m and is released from rest.The block then executes simple harmonic motion along the horizontal x-axis.What is the velocity of the block at time  ?
A) 0.82 m/s 
B) -0.82 m/s 
C) 0.30 m/s 
D) -0.30 m/s 
E) 0.00 m/s

Quizplus · Accepted Answer

To solve this problem, we can use the equations of motion for simple harmonic motion. First, we can find the angular frequency of the motion:
ω = √(k/m) = √(190 N/m / 1.6 kg) = 6.05 rad/s
Next, we can find the amplitude of the motion (the maximum displacement from equilibrium):
A = 0.080 m
Using these values, we can find the position of the block as a function of time:
x(t) = A cos(ωt)
At time t = 0, the block is at its maximum displacement to the right (positive x direction). Therefore, we can find the phase constant:
x(0) = A = 0.080 m
x(t) = 0.080 cos(6.05t)
To find the velocity of the block at time t = 0.25 s, we can take the derivative of the position function:
v(t) = dx/dt = -0.080ω sin(6.05t)
v(0.25) = -0.080ω sin(1.51) ≈ -0.822 m/s
Since the block is currently moving to the left (negative x direction), the answer is choice B, -0.82 m/s.

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