A 0.39 kg object,attached to a spring with spring constant k = 15 N/m,is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.082 m.What is its speed at the instant when its displacement is 0.041 m? (Hint: Use conservation of energy. )

A) 4.4 m/s
B) 8.8 m/s
C) 44 cm/s
D) 56.9 cm/s
E) 31.1 cm/s

Question

A 0.39 kg object,attached to a spring with spring constant k = 15 N/m,is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.082 m.What is its speed at the instant when its displacement is 0.041 m? (Hint: Use conservation of energy. )
​
A) 4.4 m/s
B) 8.8 m/s
C) 44 cm/s
D) 56.9 cm/s
E) ​31.1 cm/s

Quizplus · Accepted Answer

The conservation of energy principle states that the sum of kinetic and potential energy of a system remains constant, provided no external non-conservative forces act on the system.

At the equilibrium position (displacement = 0), all the potential energy is converted to kinetic energy, and at the displacement of 0.041 m, the object should have lost half of the potential energy at the equilibrium position.

Therefore, we can write:

(1/2) kx^2 = (1/2) mv^2

where k is the spring constant, x is the displacement, m is the mass, and v is the velocity.

Solving for v, we get:

v = sqrt((k/m) x^2)

Substituting the given values, we get:

v = sqrt((15/0.39) * (0.041)^2)

v = 0.44 m/s or 44 cm/s (rounded to one decimal place)

Therefore, the answer is C, 44 cm/s.

A 039 Kg Object,attached to a Spring with Spring Constant K