A 0.56-kg mass is suspended from a string, forming a pendulum. The period of this pendulum is 6.0 s when the amplitude is 1.5 cm. The mass of the pendulum is now reduced to 0.35 kg. What is the period of oscillation now, when the amplitude is 2.8 cm?
Since when is period dependent on mass or amplitude? Don't you have a lab to test the is exact hypothesis?
no this is a homework assignment from the textbook
There is no difference in the period because it the period is not dependent on mass or amplitude. (I just had this same problem on webassign)
To find the period of oscillation for the new setup, we can use the formula for the period of a pendulum:
T = 2π * √(L/g)
Where:
T = period (in seconds)
π = mathematical constant pi (approximately 3.14159)
L = length of the pendulum (in meters)
g = acceleration due to gravity (approximately 9.8 m/s²)
In this case, we have the following information:
Mass of the pendulum (initial): m1 = 0.56 kg
Mass of the pendulum (final): m2 = 0.35 kg
Amplitude (initial): A1 = 1.5 cm = 0.015 m
Amplitude (final): A2 = 2.8 cm = 0.028 m
Let's calculate the length of the pendulum (initial) first:
L1 = A1 / (2π)
L1 = 0.015 m / (2π)
L1 ≈ 0.0024 m
Now, let's calculate the length of the pendulum (final):
L2 = A2 / (2π)
L2 = 0.028 m / (2π)
L2 ≈ 0.0045 m
Since the length of the pendulum affects the period of oscillation, we have two different periods for the initial and final setups.
Using the formula for the period of a pendulum, let's find the initial period (T1):
T1 = 2π * √(L1/g)
T1 = 2π * √(0.0024 m / 9.8 m/s²)
T1 ≈ 2π * √0.000245 m ≈ 0.0303 s
Now, using the same formula, let's find the final period (T2):
T2 = 2π * √(L2/g)
T2 = 2π * √(0.0045 m / 9.8 m/s²)
T2 ≈ 2π * √0.000459 m ≈ 0.0341 s
Therefore, the period of oscillation for the pendulum, when the amplitude is 2.8 cm, is approximately 0.0341 seconds.