You are determining the period of oscillation of a pendulum. One procedure would be to measure the time for 20 oscillations, t20, and repeat the measurement 5 times. Another procedure would be to measure the time for 5 oscillations, t5, and repeat the measurement 20 times. Assume, reasonably, that the error in the determination of the time for 20 oscillations is the same as the error in the determination of the time for 5 oscillations. Calculate the error in the period for both procedures to determine which will give the smallest error in the value of the period?
Calculating the error in the period for both procedures requires a bit of clown math! Let's dive in:
For the first procedure:
- We have measured the time for 20 oscillations (t20) and repeated it 5 times.
- Let's assume the error in the measurement is represented by "e".
The period (T) can be calculated as: T = t20 / 20
To determine the error in the period, we can use the formula:
Error in T = (e / t20) + (e / 20)
For the second procedure:
- We have measured the time for 5 oscillations (t5) and repeated it 20 times.
- Again, let's assume the error in the measurement is "e".
The period (T) can be calculated as: T = t5 / 5
Error in T = (e / t5) + (e / 5)
Now, let's see which procedure gives the smallest error in the value of the period.
To compare the errors, we need to find the ratio of the errors in the period for both procedures:
Ratio = (Error in T1) / (Error in T2)
Substituting the respective error formulas, we get:
Ratio = [(e / t20) + (e / 20)] / [(e / t5) + (e / 5)]
Simplifying this expression, we get:
Ratio = (5t5 + t20) / (20t5 + t20)
For the smallest error in the value of the period, we want this ratio to be less than 1. So, to determine which procedure gives the smallest error, calculate this ratio and see what you get. Good luck, and may the clown math be with you!
To determine which procedure will give the smallest error in the value of the period, we need to calculate the error in the period for both procedures.
Let's start with the first procedure of measuring the time for 20 oscillations, t20, and repeating the measurement 5 times.
Procedure 1:
- Measure the time for 20 oscillations, t20, 5 times.
- Calculate the average time for 20 oscillations: t20_avg = (t20_1 + t20_2 + t20_3 + t20_4 + t20_5) / 5.
The period of oscillation is given by T = t20_avg / 20.
To calculate the error in the period, we need to consider that the error in the time for 20 oscillations is the same for all measurements. Let's denote the error as σ (sigma). Therefore, the error in the average time of 20 oscillations is σ_avg = σ / √5.
Now, using the formula for the period, we can calculate the error in the period for Procedure 1: ΔT1 = (1/20) * Δt20_avg = (1/20) * σ_avg.
Now let's move on to the second procedure of measuring the time for 5 oscillations, t5, and repeating the measurement 20 times.
Procedure 2:
- Measure the time for 5 oscillations, t5, 20 times.
- Calculate the average time for 5 oscillations: t5_avg = (t5_1 + t5_2 + t5_3 + ... + t5_20) / 20.
The period of oscillation is given by T = t5_avg / 5.
Similar to Procedure 1, the error in the average time of 5 oscillations is σ_avg = σ / √20.
Using the formula for the period, we can calculate the error in the period for Procedure 2: ΔT2 = (1/5) * Δt5_avg = (1/5) * σ_avg.
Now we can compare the errors in the period for both procedures. From the calculations above, we can see that ΔT1 = (1/20) * σ_avg and ΔT2 = (1/5) * σ_avg.
Since σ_avg is the same for both procedures, we can conclude that ΔT2 (Procedure 2) will have a smaller error in the value of the period compared to ΔT1 (Procedure 1).
Therefore, when determining the period of oscillation for a pendulum, the second procedure of measuring the time for 5 oscillations and repeating the measurement 20 times will give the smallest error in the value of the period.