A bob of mass m is suspended from a string of length L, forming a pendulum. The period of this pendulum is 1.97 s. If the pendulum bob is replaced with one of mass 1/2 m and the length of the pendulum is increased to 2.50 L, what is the period of oscillation?
The period has nothing to do with the mass.
It is proportional to the square root of the length
1.97 * sqrt(2.5)
T1 = 1.97 s.
T2 = ?.
2pi = 6.28
T1 = 2pi*sqrt(L/g).
T1^2 = 39.5*L/g,
1.97^2 = 39.5L/9.8,
L1 = 0.962 m.
L2 = 2.50 * 0.962 = 2.41 m.
T2 = 6.28*Sqrt(2.41/9.8) = 3.11s.
To find the period of oscillation for the new pendulum, we can use the formula for the period of a pendulum:
T = 2π * √(L/g)
Where T is the period of the pendulum, L is the length of the string, and g is the acceleration due to gravity.
Given that the initial pendulum has a period of 1.97 s and a length of L, we can rewrite the equation as:
1.97 = 2π * √(L/g)
Solving for g, we get:
g = (2π * √L)² / (1.97)²
Now we can find the period of the new pendulum with the mass of 1/2 m and a length of 2.50 L. Using the same formula, we get:
T' = 2π * √((2.50 L)/g')
where T' is the period of the new pendulum, L is the length of the new pendulum, and g' is the acceleration due to gravity with the new mass.
Substituting the value of g from earlier, we can calculate g':
g' = (2π * √(2.50 L))² / (1.97)²
Finally, we can substitute g' into the equation for the period of the new pendulum and solve for T':
T' = 2π * √((2.50 L)/g')
After simplifying the equation, we can evaluate T' to find the period of oscillation for the new pendulum.