Q50P
Question
Question: In the laboratory, a student studies a pendulum by graphing the angle \(\theta \) that the string makes with the vertical as a function of time \(t\), obtaining the graph shown in Fig. E14.50. (a) What are the period, frequency, angular frequency, and amplitude of the pendulum’s motion? (b) How long is the pendulum? (c) Is it possible to determine the mass of the bob?
Step-by-Step Solution
VerifiedThe time period is \(T = 1.6\,{\rm{s}}\)
The frequency of pendulum is \(0.625\,{\rm{Hz}}\)
The angular frequency of pendulum is \(3.93\,{\rm{rad/s}}\)
The amplitude of the pendulum’s motion is \({6^o}\)
A pendulum by graphing the angle \(\theta \) that the string makes with the vertical as a function of time \(t\)
The letter "\(T\) " stands for the period of time needed for the pendulum to complete one complete oscillation.
\(T = 2\pi \sqrt {\frac{L}{g}} \) …(i)
Where, \(L\) is the length of pendulum and \(g\) is the acceleration due to gravity
From the graph, the distance between two consecutive troughs or two consecutive valley will give the time period.
So, from the graph time period is \(T = 1.6\,{\rm{s}}\)
Frequency of the pendulum is expressed as,
\(f = \frac{1}{T}\)
Substitute \(T = 1.6\,{\rm{s}}\) in above equation
\(\begin{array}{c}f = \frac{1}{{1.6\,{\rm{s}}}}\\ = 0.625\,{\rm{Hz}}\end{array}\)
Hence, the frequency of pendulum is \(0.625\,{\rm{Hz}}\)
Angular frequency of the pendulum is expressed as,
\(\omega = 2\pi f\)
Substitute \(f = 0.625\,{\rm{Hz}}\) in above equation
\(\begin{array}{c}\omega = 2\pi \times 0.625\,{\rm{Hz}}\\ = 3.93\,{\rm{rad/s}}\end{array}\)
Hence, the angular frequency of pendulum is \(3.93\,{\rm{rad/s}}\)
Amplitude of the pendulum’s motion is maximum displacement in the y-axis in the graph.
From graph the amplitude of the pendulum’s motion is \({6^o}\)