A mass of 0.40 kg,hanging from a spring with a spring constant of 80 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through a point at 0.05 m displacement? The starting displacement of the mass is 0.10 m from its equilibrium position.
A)zero
B)1.4 m/s
C)1.7 m/s
D)1.2 m/s

Question

Quizplus · Accepted Answer

The total mechanical energy in simple harmonic motion is constant and is the sum of kinetic and potential energy. At maximum displacement (0.10 m), all energy is potential: $ E = \frac{1}{2} k x^2 = \frac{1}{2} \times 80 \times 0.10^2 = 0.4 \, \text{J} $. At 0.05 m displacement, the potential energy is $ \frac{1}{2} \times 80 \times 0.05^2 = 0.1 \, \text{J} $. The kinetic energy is the difference: $ 0.4 - 0.1 = 0.3 \, \text{J} $. Using $ KE = \frac{1}{2} mv^2 $, solve for $ v $: $ 0.3 = \frac{1}{2} \times 0.40 \times v^2 $, giving $ v = 1.2 \, \text{m/s} $.

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