An object with mass 2.3 kg is executing simple harmonic motion, attached to a spring with spring constant 280 N/m . When the object is 0.019 m from its equilibrium position, it is moving with a speed of 0.65 m/s. Calculate the amplitude of the motion.Calculate the maximum speed attained by the object.
from the position, calculate potential energy (1/2 k x^2), then calculate the present KE from 1/2 m v^2. Add them to get total energy. At max speed, all energy is KE, so calculate the max speed from that total energy.
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 is the angular frequency of the motion, and
A is the amplitude of the motion.
We can calculate the angular frequency using the formula:
ω = √(k / m)
Where:
k is the spring constant, and
m is the mass of the object.
Let's calculate the angular frequency first:
ω = √(280 N/m / 2.3 kg)
ω = √121.74 rad/s
ω ≈ 11.03 rad/s
Now, we can substitute the given velocity and the calculated angular frequency into the first equation to find the amplitude:
0.65 m/s = 11.03 rad/s * A
Solving for A:
A = 0.65 m/s / 11.03 rad/s
A ≈ 0.059 m
Therefore, the amplitude of the motion is approximately 0.059 meters.
To calculate the maximum speed attained by the object, we can use the equation:
v_max = ω * A
Substituting the calculated angular frequency and amplitude into the equation:
v_max = 11.03 rad/s * 0.059 m
v_max ≈ 0.651 m/s
Therefore, the maximum speed attained by the object is approximately 0.651 meters per second.