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.
To find the period of oscillation of the meter stick, we can use the formula for the period of a physical pendulum:
T = 2π√(I/((mgh)/2))
where T is the period of oscillation, I is the moment of inertia of the meter stick, m is the mass of the meter stick, g is the acceleration due to gravity, and h is the distance from the pivot point to the center of mass of the meter stick.
A) Let's find the moment of inertia of the meter stick first. The moment of inertia for a meter stick rotating about one end can be calculated as:
I = (1/3)mL^2
where m is the mass of the meter stick and L is the length of the meter stick.
B) Now let's find the distance h (the distance from the pivot point to the center of mass of the meter stick). Since the meter stick is suspended at one end, the center of mass is located at L/2, where L is the length of the meter stick.
h = L/2
C) Now we can substitute the values into the formula for the period:
T = 2π√((1/3)mL^2/((mgh)/2))
T = 2π√((1/3)Lg)
D) To calculate the period of oscillation, we need to know the length of the meter stick (L). However, in this question, the length of the light string (0.348m) is given. So, we need to convert the length of the light string to the length of the meter stick by adding the length of the meter stick's end:
L = 0.348m + 0.01m
Now we can calculate the period of oscillation using the formula:
T = 2π√((1/3)(0.348m + 0.01m) * 9.8 m/s^2)
Simplifying the equation, we get:
T ≈ 2.266 seconds (to three significant figures)
Therefore, the period of oscillation of the meter stick is approximately 2.266 seconds.
B) To find the difference in the period between the meter stick and a 0.848m long simple pendulum, we can calculate the period of a simple pendulum using the formula:
T_simple_pendulum = 2π√(L_simple_pendulum / g)
where L_simple_pendulum is the length of the simple pendulum.
Substituting the values into the formula, we get:
T_simple_pendulum = 2π√(0.848m / 9.8 m/s^2)
T_simple_pendulum ≈ 1.084 seconds (to three significant figures)
Now we can find the percentage difference between the two periods:
Percentage difference = ((T_simple_pendulum - T) / T_simple_pendulum) * 100
Percentage difference = ((1.084s - 2.266s) / 1.084s) * 100
Percentage difference ≈ -108.027% (to three significant figures)
Therefore, the period of oscillation of the meter stick differs from a 0.848m long simple pendulum by approximately -108.027%. Note that the negative sign indicates that the period of the meter stick is longer than the period of the simple pendulum.