Question

Period of a Pendulum The period of a pendulum varies directly as the square root of the length of the pendulum and inversely as the square root of the acceleration due to gravity. Find the period when the length is 121 cm and the acceleration due to gravity is 980 cm per second squared, if the period is 6π seconds when the length is 289 cm and the acceleration due to gravity is 980 cm per second squared.

Accepted Answer

Identify the relationship given: The period $$T$$ varies directly as the square root of the length $$L$$ and inversely as the square root of the acceleration due to gravity $$g. $$This can be written as the equation:

$$T = k \frac{\sqrt{L}}{\sqrt{g}}$$

where $$k is $$the constant of proportionality. Use the given values to find the constant $$k. $$Substitute $$T = 6\pi$$, $$L = 289$$, and $$g = 980$$ into the equation:

$$6\pi = k \frac{\sqrt{289}}{\sqrt{980}}$$ Solve the equation for $$k by $$isolating it on one side:

$$k = 6\pi 	imes \frac{\sqrt{980}}{\sqrt{289}}$$ Now, use the value of $$k$$ found in the previous step to find the period $$T$$ when $$L = 121$$ and $$g = 980. $$Substitute these values into the original formula:

$$T = k \frac{\sqrt{121}}{\sqrt{980}}$$ Substitute the expression for $$k$$ from step 3 into the equation in step 4 to express $$T$$ entirely in terms of known quantities, then simplify to find the period.

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Key Concepts

Direct and Inverse Variation

Square Root Function

Solving for Constants in Variation Problems