A 1024 g mass is hung on a spring. As a result the spring stretches 35 cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation? Give your answer in seconds with 3 significant figures.
I think im suppose to use T=2pi (Sqrt M/K)
Yes, you are correct. To find the period of the resulting oscillation, you can use the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.
First, let's find the spring constant (k) using the information given. The spring stretches 35 cm when the mass is hung on it. The spring constant represents the stiffness of the spring and is defined as the force required to stretch the spring by a certain amount. In this case, we need to convert the displacement from centimeters to meters, so 35 cm = 0.35 m.
Now, we can use Hooke's Law to find the spring constant. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is F = kx, where F is the restoring force, k is the spring constant, and x is the displacement.
Since the mass is at equilibrium when it stretches the spring by 35 cm, the force exerted by the spring is equal to the weight of the mass, which can be calculated using the formula F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values into Hooke's Law, we have mg = kx. Plugging in the values of the mass (1024 g = 1.024 kg) and displacement (0.35 m), we can solve for k:
k = mg/x = (1.024 kg)(9.8 m/s^2)/(0.35 m) = 28.454 N/m.
Now that we have the spring constant (k), we can calculate the period (T). Substituting the values into the formula T = 2π√(m/k), we get:
T = 2π√(m/k) = 2π√(1.024 kg/28.454 N/m) = 0.724 s (rounded to 3 significant figures).
Therefore, the period of the resulting oscillation is approximately 0.724 seconds.