A 0.56-kg mass is suspended from a string, forming a pendulum. The period of this pendulum is 3.6 s when the amplitude is 2.2 cm. The mass of the pendulum is now reduced to 0.27 kg. What is the period of oscillation now, when the amplitude is 2.5 cm? Give answer in seconds.
I was thinking that T is not dependant upon mass. I was using the equation T= (2*pi)*( sqrt L/g) =
(2pi)*(sqrt2.5/9.8) = 3.17 which is wrong.
To correctly calculate the period of oscillation for the given scenario, we need to take into account the change in mass and amplitude.
The formula you mentioned, T = 2π√(L/g), is the correct formula for the period of a simple pendulum, where T represents the period, L is the length of the pendulum, and g is the acceleration due to gravity. However, this formula assumes that the mass of the pendulum does not affect the period.
Let's first calculate the initial length of the pendulum when the amplitude is 2.2 cm. The amplitude is half the length of the swing, so the length of the pendulum will be 2 times the amplitude. Therefore, L = 2 × 2.2 cm = 4.4 cm = 0.044 m.
Now we can calculate the initial period of oscillation using the formula: T_1 = 2π√(L/g). Plugging in the values, we have: T_1 = 2π√(0.044/9.8) ≈ 1.014 seconds.
Next, let's calculate the new length of the pendulum when the amplitude is 2.5 cm. Following the same logic as before, the length of the pendulum will be 2 times the amplitude, so L = 2 × 2.5 cm = 5 cm = 0.05 m.
Finally, we can find the new period of oscillation using the formula: T_2 = 2π√(L/g). Plugging in the values, we get: T_2 = 2π√(0.05/9.8) ≈ 1.113 seconds.
Therefore, the period of oscillation for the reduced mass (0.27 kg) with an amplitude of 2.5 cm is approximately 1.113 seconds.