Question

A 128 g ball is tied to a string. It is pulled to an angle of 3.60 degrees and released to swing as a pendulum. A student with a stopwatch finds that 17 oscillations take 15.5 s .

Part A
How long is the string?

Answer 1

Concepts and reason

The concepts required to solve this problem are time period and frequency of the simple pendulum.

First, determine the frequency of pendulum by using the number of cycles per second. Then, determine the period of oscillation by using the frequency of pendulum. Finally, determine the length of the string by using the relation between the period and length of the string.

Fundamentals

Frequency is defined as the number of cycles per second. The frequency can be calculated by using the expression:

$f = \frac{N}{t}$

Here, N is the number of cycles in time t.

The period of oscillation is defined as the time taken to complete one oscillation. The period of oscillation is calculated by using the formula,

$T = \frac{1}{f}$

Here, f is the frequency.

The period of simple pendulum is given by the relation,

$T = 2\pi \sqrt {\frac{l}{g}}$

Here, l is the length of the pendulum and g is acceleration due gravity.

Calculate the period of pendulum.

The frequency of pendulum is given as,

$f = \frac{N}{t}$

Here, N is the number of cycles completed by pendulum in time t.

The period of pendulum is given as,

$T = \frac{1}{f}$

Substitute $\frac{N}{t}$ for f in the equation $T = \frac{1}{f}$.

$\begin{array}{c}\\T = \frac{1}{{\frac{N}{t}}}\\\\ = \frac{t}{N}\\\end{array}$

Substitute 15.5 s for t and 17 for N in the equation $T = \frac{t}{N}$.

$\begin{array}{c}\\T = \frac{{15.5\;{\rm{s}}}}{{17}}\\\\ = 0.9118{\rm{ s}}\\\end{array}$

Rearrange the above equation $T = 2\pi \sqrt {\frac{l}{g}}$ for l.

$\begin{array}{c}\\T = 2\pi \sqrt {\frac{l}{g}} \\\\{T^2} = 4{\pi ^2}\frac{l}{g}\\\\l = \frac{{{T^2}g}}{{4{\pi ^2}}}\\\end{array}$

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g and 0.9118 s for T in the equation $l = \frac{{{T^2}g}}{{4{\pi ^2}}}$.

$\begin{array}{c}\\l = \frac{{{{\left( {0.9118{\rm{ s}}} \right)}^2}\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{4{\pi ^2}}}\\\\ = 0.206{\rm{ m}}\left( {\frac{{100{\rm{ cm}}}}{{1{\rm{ m}}}}} \right)\\\\ = 20.6{\rm{ cm}}\\\end{array}$

Ans:

The length of the string is 20.6 cm.