An object with mass 5.0 kg is attached to a spring with spring stiffness constant 192 N/m and is executing simple harmonic motion. When the object is 0.35 m from its equilibrium position, it is moving with a speed of 0.02 m/s. Calculate the amplitude of the motion.
To calculate the amplitude of the motion, we can use the equation for the velocity of an object undergoing simple harmonic motion:
v = ω * A
where:
- v is the velocity of the object,
- ω (omega) is the angular frequency of the motion, and
- A is the amplitude of the motion.
First, we need to find the angular frequency ω. The angular frequency is related to the spring stiffness constant k and the mass m of the object by the equation:
ω = sqrt(k / m)
Given:
- k = 192 N/m (spring stiffness constant)
- m = 5.0 kg (mass of the object)
Plugging in these values, we can calculate ω:
ω = sqrt(192 N/m / 5.0 kg)
= sqrt(38.4 N/kg)
≈ 6.2 rad/s
Next, we can use the given information about the velocity v when the object is 0.35 m from its equilibrium position. At this point, the object is at its maximum displacement, so the velocity v is at its maximum value. Therefore, we can write:
v = ω * A
Given:
- v = 0.02 m/s (velocity of the object)
- ω = 6.2 rad/s (angular frequency)
Plugging in these values, we can solve for A (the amplitude):
0.02 m/s = 6.2 rad/s * A
Rearranging the equation:
A = 0.02 m/s / 6.2 rad/s
≈ 0.0032 m
Therefore, the amplitude of the motion is approximately 0.0032 m.