An experiment is carried out to measure the spring constant of a spring. A mass of 500 g is suspended on the spring. It is pulled down a small distance and the time for 20 oscillations is measured to be 34 s. a) explain why the mass performs simple harmonic motion. b) What is the spring constant? c) what is the equilibrium extension of the spring? d) if the mass and spring were to be moved to the surface of the moon (where the gravitational field strength is 1.6N/kg) what would the effect be on the time period of oscillation and on the equilibrium extension of the spring.
B)T=6.82
I will be happy to check your thinking on this. It is a straightforward problem.
answers
mass is released
a) The mass performs simple harmonic motion because it is attached to a spring and is pulled down a small distance. In simple harmonic motion, the object experiences a restoring force that is directly proportional to its displacement from the equilibrium position. The force exerted by the spring provides this restoring force, always trying to bring the mass back to its equilibrium position.
b) To determine the spring constant (k), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension. Mathematically, this can be expressed as F = kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the mass is 500 g (or 0.5 kg), and we have the time period for 20 oscillations, which is 34 s. The time period (T) of one oscillation can be found by dividing the total time by the number of oscillations, so T = 34 s / 20 = 1.7 s.
The formula for the time period of an object in simple harmonic motion is T = 2π√(m/k), where m is the mass and k is the spring constant.
We can rearrange this formula to solve for k: k = (4π²m) / T². Plugging in the values, we get k = (4π² * 0.5) / (1.7²) = 3.297 N/m.
Therefore, the spring constant is 3.297 N/m.
c) The equilibrium extension of the spring can be determined by analyzing the system when there are no external forces acting on it. At equilibrium, the force exerted by the spring is equal in magnitude and opposite in direction to the gravitational force on the mass. This means that the weight of the mass (mg) is balanced by the force exerted by the spring (kx), so mg = kx.
Rearranging this equation, we can solve for x: x = mg/k. Plugging in the values, we get x = (0.5 * 9.8) / 3.297 = 1.51 m.
Therefore, the equilibrium extension of the spring is 1.51 m.
d) If the mass and spring were moved to the surface of the moon, where the gravitational field strength is 1.6 N/kg, the effect on the time period of oscillation and the equilibrium extension of the spring can be determined.
The time period (T) of oscillation is inversely proportional to the square root of the gravitational field strength, so T ∝ 1/√g.
On the moon, g = 1.6 N/kg. Therefore, the new time period (T_moon) would be T_moon ∝ 1/√1.6 = 0.7906 T.
Thus, the time period of oscillation would decrease if the mass and spring were moved to the moon's surface.
The equilibrium extension of the spring, however, would remain the same. The extension of the spring depends on the mass and the spring constant, not on the gravitational field strength.
Therefore, the equilibrium extension of the spring would not be affected by the change in gravitational field strength.