Question

TUDIEM 14 Geoff counts the number of oscillations of a simple pendulum at a location where the acceleration due to gravity is 9.80 m/s, and finds that it takes 28.0 s for 17 complete cycles. 1) Calculate the length of the pendulum. (Express your answer to three significant figures.) m Submit

Answer 1

Solution)

Given,

Number of revolutions, n=17

Time, t=28 sec

So, Time Period, T=28/17= 1.64 s

We know, T=2pi*root(l/g)

So, length, L= T^2*g/4pi^2

Substitute values,

L=(1.64)^2*9.8/(4pi^2)= 0.668 m(Ans)

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Answer 2

A simple pendulum's period depends on its length and acceleration due to gravity and is independent of the pendulum's mass. Therefore, near the surface of Earth, the only quantity that separates one pendulum from another is the length of the connector between its axis of rotation and the pendulum bob. The longer the connector, the longer the pendulum's period.

The period ?$T$ of a pendulum is the time for one complete cycle of an oscillation. Thus, given the time ?$t$ for a given number ?$n$ of cycles, the period is the ratio of ?$t$ to ?.$n\text{.}$

$$T=\frac{t}{n}$$

The period of a pendulum is determined by the pendulum's length ?$L$ and the acceleration due to gravity ?.$g\text{.}$

$$T=2\mathit{\pi }\sqrt{\frac{L}{g}}$$

Substitute the expression for ?,$T\text{,}$ then solve for ?.$L\text{.}$

$$\begin{array}{rl}{\displaystyle \frac{t}{n}}& =2\mathit{\pi }\sqrt{{\displaystyle \frac{L}{g}}}\\ \\ \sqrt{{\displaystyle \frac{L}{g}}}& ={\displaystyle \frac{t}{2\mathit{\pi }n}}\\ \\ L& ={\left(\frac{t}{2\mathit{\pi }n}\right)}^{2}g\end{array}$$

Substitute the given values and evaluate.

$$\begin{array}{rl}L& ={\left[\frac{8.85\text{ s}}{2\mathit{\pi }\left(14.0\right)}\right]}^2\left(9.80{\text{ m/s}}^2\right)\\ \\ & =0.0992\text{ m}\end{array}$$

Therefore, the pendulum is 0.0992 m$0.0992\text{ m}$ long