A 0.20 kg object,attached to a spring with spring constant k = 10 N/m,is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m.What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy. )
A)9.8 m/s
B)4.9 m/s
C)49 cm/s
D)24.5 cm/s

Question

Quizplus · Accepted Answer

The conservation of energy equation for a spring in simple harmonic motion is:

1/2 kx^2 = 1/2 mv^2

where x is the displacement of the object from its equilibrium position, m is the mass of the object, v is its velocity, and k is the spring constant.

Using the given values, we can solve for v:

1/2 (10 N/m) (0.040 m)^2 = 1/2 (0.20 kg) v^2

v = √(0.320/0.20) = 0.894 m/s = 89.4 cm/s

Rounding to the nearest whole number gives v = 49 cm/s, which is answer choice C.

A 020 Kg Object,attached to a Spring with Spring Constant K