A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 1.1 m/s. Determine the amplitude A of the motion.
To determine the amplitude A of the motion, we need to utilize the relationship between the maximum speed and the amplitude for an object undergoing simple harmonic motion.
In simple harmonic motion, the maximum speed occurs at the equilibrium position, where the spring is neither compressed nor extended. At this point, all the potential energy is converted into kinetic energy, resulting in the maximum speed.
The relationship between the maximum speed (v_max) and the amplitude (A) for simple harmonic motion is given by the equation:
v_max = ωA
where ω is the angular frequency of the motion.
To find the angular frequency (ω), we can use the relationship between the spring constant (k) and the angular frequency:
ω = sqrt(k / m)
where m is the mass of the object.
Let's solve step by step:
1. Convert the given mass to kilograms:
mass = 0.20 kg
2. Calculate the angular frequency using the spring constant and mass:
k = 120 N/m
ω = sqrt(k / m)
= sqrt(120 / 0.20)
= sqrt(600)
≈ 24.49 rad/s
3. Substitute the angular frequency and maximum speed into the equation to solve for the amplitude:
v_max = ωA
1.1 m/s = 24.49 rad/s * A
4. Rearrange the equation to solve for A:
A = v_max / ω
= 1.1 m/s / 24.49 rad/s
≈ 0.045 m
Therefore, the amplitude (A) of the motion is approximately 0.045 m.