A meter stick, suspended at one end by a 0.348m long light string, is set into oscillation. The acceleration of gravity is 9.8 m/s^2.
A)Determine the period of oscillation.
Answer in units of s.
B)By what percentage does this differ from a 0.848m long simple pendulum?
Answer in units of percent.
Moment of inertia about suspension point = I about stick middle + mass*(.348+.5)^2
=(1/12) m (1)^2 + m (.719)
= .802 m
T = 2 pi sqrt (I/mgL) = 2 pi sqrt(.802m/(9.8m*.848) )
= 2 pi sqrt(.0965)
= 1.95 seconds
simple pendulum length .848
T = 2 pi sqrt (.848/9.8)
= 1.84 seconds
you can do he percent. Check my arithmetic!
To determine the period of oscillation of the meter stick, we can use the formula for the period of a simple pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
A) We are given that the length of the string is 0.348 m and the acceleration due to gravity is 9.8 m/s^2. Substituting these values into the formula, we have:
T = 2π√(0.348/9.8)
Now we can calculate the value of T:
T = 2π√(0.0355)
T ≈ 2π * 0.1887
T ≈ 1.183 s
So the period of oscillation of the meter stick is approximately 1.183 seconds.
B) To determine the difference in percentage between the period of the meter stick and a 0.848 m long simple pendulum, we can use the formula:
% Difference = ((T1 - T2) / T2) * 100
where T1 is the period of the meter stick and T2 is the period of the simple pendulum.
We already calculated T1 as 1.183 s. Now we need to calculate T2 for the simple pendulum using the same formula as before:
T2 = 2π√(0.848/9.8)
T2 = 2π√(0.0865)
T2 ≈ 2π * 0.2944
T2 ≈ 1.849 s
Now we can calculate the percentage difference:
% Difference = ((1.183 - 1.849) / 1.849) * 100
% Difference ≈ (-0.666 / 1.849) * 100
% Difference ≈ -0.360 * 100
% Difference ≈ -36 %
Therefore, the period of oscillation of the meter stick differs from that of a 0.848 m long simple pendulum by approximately 36 percent.