A 215 g mass is attached to a spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the system is 2.30 J.
(a) Find the force constant of the spring.
135.81 N/m
(b) Find the amplitude of the motion.
i just need help with what equation i should use. im stumped on this question.
The period is
P = 2 pi sqrt(k/m)
Use that and the mass to get the spring constant, k.
The total energy of the system is
(1/2) k X^2 = 2.30 J,
where X is the amplitude. .
Knowing that and k, solve for X.
To find the force constant of the spring and the amplitude of the motion, we can use the formulas related to simple harmonic motion and energy.
(a) Force constant (k):
The formula for the period (T) of an object undergoing simple harmonic motion is given by:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the force constant.
To find the force constant, we need to rearrange the formula:
k = (2π)^2(m/T^2)
where T = 0.250 s and m = 215 g. We need to convert the mass from grams to kilograms by dividing it by 1000 (1 kg = 1000 g).
k = (2π)^2(0.215 kg) / (0.250 s)^2
k = 135.81 N/m
Therefore, the force constant of the spring is 135.81 N/m.
(b) Amplitude (A):
The total energy (E) of the system is given by:
E = (1/2)kA^2
where E is the total energy and A is the amplitude of the motion.
To find the amplitude, we can rearrange the formula:
A = √(2E/k)
where E = 2.30 J and k = 135.81 N/m.
A = √(2 * 2.30 J / 135.81 N/m)
A = √(0.0337 m^2)
A = 0.1835 m
Therefore, the amplitude of the motion is 0.1835 m.