Physics > Simple Harmonic Motion > 10.0 Method for Calculating Time Period of a Simple Harmonic Motion
Simple Harmonic Motion
1.0 Types of Motion
2.0 Causes of Oscillation
3.0 Solution of the Equation of SHM
4.0 Kinematics of SHM equation i.e., $x = A\sin \left( {\omega t + \phi } \right)$
5.0 Relation between Simple Harmonic Motion and Uniform Circular Motion
6.0 Sign Convention of a Simple Harmonic Motion
7.0 How to Write the Simple Harmonic Motion Equation
8.0 Force and Energy in Simple Harmonic Motion
9.0 Basic Differential Equation of SHM
10.0 Method for Calculating Time Period of a Simple Harmonic Motion
10.1 Restoring Force or Torque Method
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
11.0 Spring Block System
12.0 Physical Pendulum
13.0 Vector Method of Combining Two or More Simple Harmonic Motions
14.0 Simple Harmonic Oscillation of a Fluid Column
10.3 Important points
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
Following points should be remembered in case of a simple pendulum:
1. For large amplitudes the approximations $\sin \theta \approx \theta $ is not valid and the calculations of the period is more complex.
The time period in this case depends on the amplitude ${\theta _o}$ and is given by, $$T = 2\pi \sqrt {\frac{l}{g}\left( {1 + \frac{{\theta _o^2}}{{16}}} \right)} $$
Here the amplitude ${\theta _o}$ must be expressed in radians. This approximation is sufficient for most practical situations.
2. If the time period of a simple pendulum is $2$ seconds, it is called as seconds pendulum.
3. If length of the pendulum is large, $g$ no longer remain vertical but will be directed towards the center of the earth and expression for time period is given by, $$T = 2\pi \sqrt {\frac{1}{{g\left( {\frac{1}{l} + \frac{1}{R}} \right)}}} $$
Here $R$ is the radius of the earth. From this expression we can see that,
- If $l \ll R,\frac{1}{l} \gg \frac{1}{R}$, then $T = 2\pi \sqrt {\frac{l}{g}} $
- As $l \to \infty ,\frac{1}{l} \to 0andT = 2\pi \sqrt {\frac{R}{g}} $, and substituting the value of $R$ and $g$ we get $T=84.6$ minutes.
4. Due to change in temperature, length of pendulum and so the time period will change.
If $\Delta \theta $ is the increase in temperature then,
$$\begin{equation} \begin{aligned} l' = l\left( {1 + \alpha \Delta \theta } \right)\ \ or\ \ \frac{{l'}}{l} = 1 + \alpha \Delta \theta \\ \frac{{T'}}{T} = \sqrt {\frac{{l'}}{l}} = {\left( {1 + \alpha \Delta \theta } \right)^{\frac{1}{2}}}\;\quad \left( {sinceT \propto \sqrt l } \right) \\ \frac{{T'}}{T} \approx \left( {1 + \frac{1}{2}\alpha \Delta \theta } \right) \\ \frac{{T'}}{T} - 1 = \frac{1}{2}\alpha \Delta \theta \\ \frac{{T' - T}}{T} = \frac{1}{2}\alpha \Delta \theta \\ \Delta T = \frac{1}{2}T\alpha \Delta \theta \\\end{aligned} \end{equation} $$
Note: In case of a pendulum clock, time is lost if $T$ increases and gained if $T$ decreases.
The gain or loss in time $t$ is given by,
$$\Delta t = \frac{{\Delta T}}{{T'}}.t$$
For example, if $T=2s$, $T'=3s$, then $\Delta T = 1\sec $
Therefore, time lost by clock in $1$ hour is $$\Delta t = \frac{1}{3}{\text{}}.{\text{}}3600 = 1200{\text{s}}$$
5. Simple Harmonic Motion of a Conical Pendulum
As observed from Fig. SHM 18, $\theta$ is constant, bob describes horizontal circle,
$$\begin{equation} \begin{aligned} Tcos\theta = mg \\ \frac{T}{m} = \frac{{mg}}{{cos\theta }} \\ {g_{eff}} = \frac{g}{{cos\theta }} \\\end{aligned} \end{equation} $$
As we know that, $$T = 2\pi \sqrt {\frac{l}{{{g_{eff}}}}} $$ Therefore, $$T = 2\pi \sqrt {\frac{{lcos\theta }}{g}} \quad or\quad 2\pi \sqrt {\frac{h}{g}} $$
Example 4. A simple pendulum is executing simple harmonic motion with a time period $T$. If the length of the pendulum is increased by $44\%$. Find the percentage increase in the time period.
Solution: The time period of simple pendulum is, $$T = 2\pi \sqrt {\frac{l}{g}} $$
Since the length is increased by $44\%$, so the new length becomes $$l' = 1.44l$$
So, the new time period, $$\begin{equation} \begin{aligned} T' = 2\pi \sqrt {\frac{{1.44l}}{g}} = 1.2\left( {2\pi \sqrt {\frac{l}{g}} } \right) \\ T' = 1.2T \\\end{aligned} \end{equation} $$
Change in the time period is $$T' - T = 1.2T - T = 0.2T$$
Percentage change in the time period is $$\frac{{0.2T}}{T} \times 100 = 20\% $$
