Question

28. (a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79m/s2 is moved to a location where it the acceleration due to gravity is 9.82m/s2. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.
The answer is "(a) 2.99541 s(b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by (0.01)2=0.01% so it is necessary to have at least 4 digits after the decimal to see the changes." But please include the steps to each solution.

Answer 1

Answer:

See below ↓

Explanation:

Assuming the motion of a simple pendulum, the time period is given by

• ⇒ T = 2π√(L/g)
• ⇒ T₁/T₂ = √(g₂/g₁)

The values we have

• T₁ = 3 s
• g₁ = 9.79 m/s²
• g₂ = 9.82 m/s²

Solving

• T₂ = T₁/√(g₂/g₁)
• T₂ = 3/√(9.82/9.79)
• T₂ = 3.004593 seconds

• This many digits are needed because the acceleration is changed by a very small percentage (roughly 2 decimal places)
• So on taking the square root of the ratio of the gravity due to acceleration (2 values), 4 decimal places will be present at the very least