Pendulum A with mass m and length l has a period of T. If pendulum B has a mass of 2m and a length of 2l, how does the period of pendulum B compare to the period of pendulum A?
A. The period of pendulum B is 2 times that of pendulum A.
B. The period of pendulum B is half of that of pendulum A.
C. The period of pendulum B is 1.4 times that of pendulum A.
D. The period of pendulum B is 0.71 times that of pendulum A.
E. The period of pendulum B is the same as that of pendulum A.
T^2 is proportional to L/g. The mass has no effect.
So, if A^2 = L/g
B^2 = 2L/g
B^2/A^2 = 2
B/A = √2 ≈ 1.4
answer C
The period of a simple pendulum is given by the formula T = 2π√(l/g), where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity.
For pendulum A with length l, the period is T.
For pendulum B with length 2l, the period can be calculated using the same formula as T' = 2π√(2l/g) = √2 * (2π√(l/g)) = √2 * T.
Therefore, the period of pendulum B is √2 times that of pendulum A.
So the correct answer is:
C. The period of pendulum B is 1.4 times that of pendulum A.
To determine how the period of pendulum B compares to the period of pendulum A, we can use the formula for the period of a pendulum:
T = 2π√(l/g)
where T is the period, l is the length of the pendulum, and g is acceleration due to gravity (approximately 9.8 m/s^2).
Pendulum A has a mass of m and length l, so its period is T.
Pendulum B has a mass of 2m and length 2l. To find its period, we substitute these values into the formula:
Tb = 2π√(2l/g) = 2π√2(√l/g) = 2√2 × T
Therefore, the period of pendulum B is 2√2 times (approximately 2.83 times) the period of pendulum A.
Comparing this to the options given:
A. The period of pendulum B is 2 times that of pendulum A. - Not correct.
B. The period of pendulum B is half of that of pendulum A. - Not correct.
C. The period of pendulum B is 1.4 times that of pendulum A. - Not correct.
D. The period of pendulum B is 0.71 times that of pendulum A. - Not correct.
E. The period of pendulum B is the same as that of pendulum A. - Not correct.
Therefore, the correct answer is that the period of pendulum B is approximately 2.83 times that of pendulum A.