A 50 gram mass vibrates in SHM at the end of a spring. The amplitude of the motion is
0.12 meters, and the period is T= 1.70 seconds.
Find:
The maximum acceleration of the mass?
The speed of the mass when the displacement is 0.06 meters?
To find the maximum acceleration of the mass in Simple Harmonic Motion (SHM), we can use the formula:
a_max = (2π/T)^2 * x_max
Where:
a_max is the maximum acceleration
T is the period of the motion
x_max is the amplitude of the motion
In this case, we are given:
T = 1.70 seconds
x_max = 0.12 meters
Plugging these values into the formula, we get:
a_max = (2π/1.70)^2 * 0.12
Evaluating this expression, we get:
a_max ≈ 11.67 m/s^2
Therefore, the maximum acceleration of the mass is approximately 11.67 m/s^2.
Now, to find the speed of the mass when the displacement is 0.06 meters, we can use the formula for the velocity in SHM:
v = 2π * sqrt((k/m) * (x_max^2 - x^2))
Where:
v is the velocity
k is the spring constant
m is the mass
x is the displacement (in this case, 0.06 meters)
To find the spring constant (k), we need additional information about the system. If the system is ideal, such as a simple mass-spring system, we can use the formula:
k = (2π/T)^2 * m
Substituting the given values, we have:
k = (2π/1.70)^2 * 0.050
Evaluating this expression, we get:
k ≈ 4.347 N/m (Newtons per meter)
Now we can calculate the speed. Substituting the values into the velocity formula, we get:
v = 2π * sqrt((4.347/0.050) * (0.12^2 - 0.06^2))
Calculating this expression, we find:
v ≈ 2.73 m/s
Therefore, the speed of the mass when the displacement is 0.06 meters is approximately 2.73 m/s.