A cart of mass 223 g is placed on a frictionless horizontal air track. A spring having a spring constant of 10.20 N/ m is attached between the cart and the left end of the track. When in equilibrium, the cart is located 11.0 cm from the left end of the track. If the cart is displaced 4.20 cm from its equilibrium position, find
(a) the period at which it oscillates
T=2•π•sqrt(m/k)= 2•π•sqrt(0.223/10.20)= …
To find the period of oscillation of the cart, we can use Hooke's law and Newton's second law.
Hooke's law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position:
F = -kx
where F is the force, k is the spring constant, and x is the displacement.
In this case, the cart is displaced 4.20 cm from its equilibrium position, so x = 0.0420 m.
Using Newton's second law, we can relate the force exerted by the spring to the acceleration of the cart:
F = ma
Since the cart is on a frictionless horizontal air track, the only force acting on it is the force exerted by the spring. Therefore, F = -kx.
Plugging this into Newton's second law, we have:
-kx = ma
Rearranging the equation, we get:
a = -(k/m)x
The equation of motion for simple harmonic motion is:
a = -ω^2x
Comparing this equation with the equation we derived, we can see that ω^2 = (k/m).
The period of oscillation, T, is related to the angular frequency, ω, by the equation:
T = 2π/ω
Substituting ω^2 = (k/m) into the equation for T, we get:
T = 2π√(m/k)
Now we can find the period of oscillation.
Given that:
m = 223 g = 0.223 kg
k = 10.20 N/m
Substituting these values into the equation for T, we have:
T = 2π√(0.223/10.20)
Calculating this using a calculator, we get:
T ≈ 0.560 seconds
Therefore, the period at which the cart oscillates is approximately 0.560 seconds.