a student standing in an elevator at rest notices that her pendulum has a frequency of oscillation of 0.5 Hz.
a. what is the length of the pendulum's string?
b. the elevator starts to accelerate upwards. what effect will this have on the oscillation? explain.
To find the length of the pendulum's string, we can use the formula for the frequency of a pendulum:
f = 1 / (2π√(L/g))
Where:
f = frequency of oscillation (0.5 Hz)
L = length of the pendulum's string (unknown)
g = acceleration due to gravity (9.8 m/s^2)
Now, let's solve for L:
0.5 = 1 / (2π√(L/9.8))
To make it easier, let's isolate the variable L:
√(L/9.8) = 1 / (2π * 0.5)
Now, square both sides of the equation to remove the square root:
L/9.8 = (1 / (2π * 0.5))^2
L = (1 / (2π * 0.5))^2 * 9.8
Now, calculate the value of L using a calculator:
L ≈ 9.8 * 0.25 / (2π)^2
L ≈ 9.8 * 0.25 / 39.48
L ≈ 0.0625 / 39.48
L ≈ 0.00158 meters (or 1.58 cm)
So, the length of the pendulum's string is approximately 0.00158 meters (or 1.58 cm).
Now, let's analyze the effect of the upward acceleration of the elevator on the pendulum's oscillation. When the elevator starts to accelerate upwards, the student inside will experience a pseudo-force in the opposite direction of the elevator's motion. This pseudo-force is similar to the force of gravity, except it acts in the opposite direction.
This pseudo-force will make the effective value of gravity in the elevator less than 9.8 m/s^2. As a result, the period (or time taken for one complete oscillation) of the pendulum will increase.
In simple terms, the pendulum will take longer to complete each swing. The frequency of oscillation will decrease. This is because the apparent weight of the pendulum decreases when the upward acceleration of the elevator counteracts a portion of the real gravitational force acting on it.
So, the effect of the upward acceleration of the elevator will cause the pendulum to oscillate with a lower frequency than when it was at rest.