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 x = 0.00 m 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
T = 0.40 s?
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

Quizplus · Accepted Answer

The angular frequency of the system is:
ω = √(k/m) = √(190 N/m ÷ 1.6 kg) = √118.75 rad/s

At time t=0.40 s, the block has moved a distance of:
x = A cos(ωt) = 0.080 m cos(47.48 rad/s) ≈ 0.061 m

The velocity of the block at time t=0.40 s is:
v = -Aω sin(ωt) = -0.080 m × 47.48 rad/s × sin(18.99 rad) ≈ -0.82 m/s

The negative sign indicates that the velocity is to the left (opposite to the positive x-axis). Therefore, the correct answer is (A) -0.82 m/s.

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