Question

The formula for the period of a simple pendulum is T=2x√1/y Such a pendulum is used to determine y. The functional error in the measurement of the period T=±x and that in the measurement of the length (l) is ±y. What is the functional error in the calculated value of y?

Answer 1

Question #81155, Physics / Other

The formula for the period of a simple pendulum is $T = 2 \times V / \gamma$. Such a pendulum is used to determine $\gamma$. The functional error in the measurement of the period $T = \pm x$ and that in the measurement of the length (l) is $\pm \gamma$. What is the functional error in the calculated value of $\gamma$?

Solution

$$y = \frac{4x^2}{T^2}$$ $$\frac{\partial y}{\partial x} = \frac{8x}{T^2}, \frac{\partial y}{\partial T} = -\frac{8x^2}{T^3}$$ $$\Delta y = \sqrt{\left(\frac{8x}{T^2} \delta y\right)^2 + \left(-\frac{8x^2}{T^3} \delta x\right)^2}$$

The functional error in the calculated value of $\gamma$:

$$\Delta y = \sqrt{\left(\frac{8x}{T^2} \delta y\right)^2 + \left(\frac{8x^2}{T^3} \delta x\right)^2}$$

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