An object with mass 8.2 kg is attached to a spring with spring stiffness constant 370 N/m and is executing simple harmonic motion. When the object is 0.16 m from its equilibrium position, it is moving with a speed of 0.31 m/s. Calculate the maximum velocity attained by the object.
(1/2)MV^2 + (1/2)kX^2 = Total energy
= 0.394 + 4.736 = 5.13 Joules
For the maximum velocity (when X = 0) solve
(1/2) M V^2 = 5.13 J
V = 1.119 m/s
To calculate the maximum velocity attained by the object, we need to first find the amplitude of the simple harmonic motion. The amplitude represents the maximum displacement of the object from its equilibrium position.
Given that the object is 0.16 m from its equilibrium position, this distance represents half of the amplitude. Therefore, the full amplitude can be calculated by multiplying the given distance by 2:
Amplitude = 2 * 0.16 m
Amplitude = 0.32 m
Next, we can use the amplitude to find the maximum velocity attained by the object. In simple harmonic motion, the maximum velocity occurs when the object is at the equilibrium position and moving in the opposite direction of its displacement.
The maximum velocity can be calculated using the equation:
Maximum Velocity = Amplitude * Angular Frequency
The angular frequency, denoted by ω, can be calculated using the equation:
ω = √(spring stiffness constant / mass)
Substituting the given values, we get:
ω = √(370 N/m / 8.2 kg)
ω ≈ 3.1725 rad/s (rounded to four decimal places)
Finally, we can substitute the calculated values into the equation for maximum velocity:
Maximum Velocity = 0.32 m * 3.1725 rad/s
Maximum Velocity ≈ 1.016 m/s (rounded to three decimal places)
Therefore, the maximum velocity attained by the object is approximately 1.016 m/s.