An object is attached to an ideal spring. It undergoes a simple harmonic motion with a total energy of E = 2.2 J. The amplitude of the motion is 0.7
m
and the maximum speed of the object is 0.5 m/s. Find the spring constant.
To find the spring constant, we can use the equation for the total energy of a system undergoing simple harmonic motion:
E = (1/2)kA^2
where:
E = total energy
k = spring constant
A = amplitude of motion
In this case, we are given that the total energy (E) is 2.2 J and the amplitude (A) is 0.7 m. We can plug these values into the equation and solve for the spring constant (k):
2.2 J = (1/2)k(0.7 m)^2
First, let's square the amplitude:
(0.7 m)^2 = 0.49 m^2
Now, we can simplify the equation:
2.2 J = (1/2)k(0.49 m^2)
Multiplying both sides by 2:
4.4 J = k(0.49 m^2)
Finally, isolate k by dividing both sides by 0.49 m^2:
k = 4.4 J / 0.49 m^2 ≈ 8.98 N/m
Therefore, the spring constant is approximately 8.98 N/m.