Question

A student measures the displacement from the equilibrium of a stretched spring and reports it to be 100 micro meter with a 1% error. The spring constant k is known to be 10N/m with 0.5% error. The percentage error in the estimate of the potential energy is V=(1/2) K x2 is

a. 0.8%

b. 2.5%

c. 1.5%

d. 3.0%

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Answer 1

Given info : A student measures the displacement from the equilibrium of a stretched spring and reports it to be 100 micro meter with a 1% error. The spring constant k is known to be 10N/m with 0.5% error.

To find : the percentage error in the estimate of the potential energy is v = 1/2 Kx² , is ...

solution : here, percentage error in the displacement of stretched spring = 1 %

percentage error in the spring constant = 0.5 %

we have, potential energy , U = 1/2 Kx²

take ln both sides,

lnU = ln(1/2) + lnK + 2lnx

differentiating both sides, we get,

dU//U = 0 + dK/K + 2dx/x

⇒ ΔU/U = ΔK/K + 2Δx/x

⇒ ΔU/U × 100 = ΔK/K × 100  + 2Δx/x × 100

⇒ % error in U = % error in K + 2 × % error in x

                        = 0.5 % + 2 × 1 % = 2.5 %

therefore the percentage error in the estimate of the potential energy is 2.5 %