I have no idea what equation to use to solve this, the answer was given to be 87 micro seconds, can someone please teach me how to go about solving this problem
A simple pendulum, consisting of a mass on a thin aluminum rod has a period of 1 second at 10 Celsius, to the nearest microsecond, by how much is its period increased at 17 Celsius ?
A simple pendulum, consisting of a mass on a thin aluminum rod has a period of 1 second at 10 Celsius, to the nearest microsecond, by how much is its period increased at 17 Celsius ?
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First you have to look up the coefficient of thermal expansion for aluminum, C per °Celcius
Now,
T = 2π√[L/g] (Simple pendulum)
From which we see that T varies with the √ of L. But L = L*(1+C*dT) where dT = change in temperature....In this problem dT = +7°, so L = Li*(1+7C)
Ergo, the factor F = 1+7C is the factor by which the length changes, so the effect on the period T is √(1+7C)
Now,
T = 2π√[L/g] (Simple pendulum)
From which we see that T varies with the √ of L. But L = L*(1+C*dT) where dT = change in temperature....In this problem dT = +7°, so L = Li*(1+7C)
Ergo, the factor F = 1+7C is the factor by which the length changes, so the effect on the period T is √(1+7C)
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It depends what value you use for the coefficient of expansion of aluminium. I'll use 2.4x10⁻⁵ /⁰C.
The period T is given by T=2π√(L/g). This is 1second at 10⁰C.
The temperature increases by 17-10 = 7⁰C.
The new length = L(1 + 7x2.4x10⁻⁵) = 1.000168L
The new period T' is given by
T' = 2π√(1.000168L/g)
= √(1.000168) x 2π√(L/g)
= 1.000084 x 1
= 1.000084s
So the period has increased from 1s to 1.000084s, which is 84μs.
The period T is given by T=2π√(L/g). This is 1second at 10⁰C.
The temperature increases by 17-10 = 7⁰C.
The new length = L(1 + 7x2.4x10⁻⁵) = 1.000168L
The new period T' is given by
T' = 2π√(1.000168L/g)
= √(1.000168) x 2π√(L/g)
= 1.000084 x 1
= 1.000084s
So the period has increased from 1s to 1.000084s, which is 84μs.