Suppose a simple pendulum is used to measure the acceleration due to gravity at various points on the Earth. If g varies by 0.151 % over the various locations where it is sampled, what is the corresponding variation in the period of the pendulum? Assume that the length of the pendulum does not change from one site to another
deltaT=sqrt(deltag/g)=sqrt(.00151)
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To find the corresponding variation in the period of the pendulum due to the variation in the acceleration due to gravity (g), 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.
We are given that g varies by 0.151% over the various locations. Let's call this variation Δg.
Δg = 0.151% = 0.151/100 = 0.00151 (decimal form)
Now, let's find the corresponding variation in the period of the pendulum, which we'll call ΔT.
ΔT = 2π * √(L/g2) - 2π * √(L/g1)
Where g1 is the initial value of g and g2 is the new value of g.
Since we are assuming that the length of the pendulum remains the same, we can cancel out the L term:
ΔT = 2π * (√1/g2 - √1/g1)
Now, substituting the values:
ΔT = 2π * (√1/(1 + Δg) - √1/1)
ΔT = 2π * (√1/(1 + 0.00151) - 1)
Now, we can use a calculator to compute this value:
ΔT ≈ 2π * (√1/1.00151 - 1) ≈ 2π * (√0.99849 - 1) ≈ 2π * (0.999245 - 1)
ΔT ≈ 2π * (-0.000755) ≈ -0.004745
Therefore, the corresponding variation in the period of the pendulum is approximately -0.004745 seconds. Note that the negative sign indicates a decrease in the period.