A 50-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 15m/s2 and its maximum speed is 3.5m/s .
To find the spring constant (k) and the amplitude (A) of the motion, we can use the given information about the maximum acceleration and maximum speed.
The maximum acceleration (amax) of the simple harmonic motion is given as 15 m/s^2. This acceleration occurs when the displacement is at the maximum positive or negative amplitude, meaning the mass is at the maximum elongation or compression from the equilibrium position of the spring.
Using the equation of motion for simple harmonic motion:
amax = ω^2 * A
Where:
amax is the maximum acceleration,
ω is the angular frequency (ω = 2πf, where f is the frequency),
A is the amplitude of the motion.
We can rearrange this equation to solve for the amplitude:
A = amax / ω^2
Now, let's find the frequency (f) using the given maximum speed (vmax) of 3.5 m/s. At the maximum velocity, the displacement from equilibrium is zero, meaning the mass is passing through the equilibrium position.
Using the equation of motion for simple harmonic motion:
vmax = ω * A
We can rearrange this equation to solve for the frequency:
f = vmax / A
First, we need to calculate the frequency (f):
f = 3.5 m/s / A
Now, we can substitute this value of frequency into the equation to find the amplitude (A):
A = amax / (2πf)
Given:
amax = 15 m/s^2
vmax = 3.5 m/s
Substituting the values:
f = 3.5 m/s / A
A = 15 m/s^2 / (2π * (3.5 m/s / A))
We can solve this equation for A:
A = 0.85 m
Now, we can find the frequency:
f = 3.5 m/s / A
f = 3.5 m/s / 0.85 m
Simplifying, we find:
f = 4.118 Hz
Finally, with the frequency known, we can calculate the spring constant (k) using the equation:
k = (2πf)^2 * m
Given:
m = 50 g = 0.05 kg
f = 4.118 Hz
Substituting the values:
k = (2π * (4.118 Hz))^2 * 0.05 kg
Simplifying, we find:
k ≈ 52.94 N/m
Therefore, the spring constant (k) is approximately 52.94 N/m and the amplitude (A) is approximately 0.85 m.