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Time Period of Simple Pendulum

Q: A simple pendulum has a time period \$T_1\$ when on the earth's surface and \$T_2\$ when taken to a height \$R\$ above the earth's surface, where \$R\$ is the radius of the earth. The value of \$\dfrac{T_2}{T_1}\$ is:

Time period of a simple pendulum \$=2\pi\sqrt{\dfrac{l}{g}}\$\$\Rightarrow g\$ on earth's surface \$=\dfrac{GM}{R^2}\$ \$g\$ on a height \$R\$ above earth's surface \$=\dfrac{GM}{\left(2R\right)^2}\$\$\Rightarrow T_1\$ (time period on earth's surface ) \$=2\pi \sqrt{\dfrac{l\left(R^2\right)}{GM}}\$ \$T_2\$ (time period at a height \$'h'\$ above earth's surface ) \$=2\pi \sqrt{\dfrac{l\left(2R\right)^2}{GM}}\$\$\Rightarrow \dfrac{T_2}{T_1}=\sqrt{4}=2.\$Hence, the answer is \$2.\$

Q: What is a simple pendulum? Show that the motion of the pendulum is SHM and hence deduce an expression for the time period of pendulum. Also, define Second's pendulum.

Simple pendulum is a point mass body suspended by a weightless thread or string from a rigid support about which it is free to oscillate.Now restoring force acting on body can be given byF = -mgsin\$\theta\$ (from figure)If \$\theta\$ is smallF = -mg\$\theta\$ ( \$\because sin \theta \approx \theta\$)Let P be any point an the path of pendulum and makes \$\angle OMP = \theta\$ which is small and arc OP be x (displacemebt) then\$\theta = \frac{OP}{1} = \frac{X}{1}\$\$\Rightarrow F = \frac{-mgx}{1}\$ ........(1)\$\Rightarrow\$ force \$\propto\$ displacement and opposite in direction hence an SHMNow this of form F = -kx...............(2)\$\Rightarrow\$ k = mg/I (from (1) and (2)\$T = 2\pi \sqrt{\dfrac{inertial\ factor}{spring\ factor}}\$\$T = 2\pi \sqrt{\dfrac{m}{mg/1}}\$\$\Rightarrow T = 2\pi \sqrt{\dfrac{1}{g}}\$

Q: The length of the simple pendulum which ticks seconds is:

The time period of a simple pendulum is\$T = 2 \pi \sqrt{\dfrac{L}{g}}\$where L is the length of the pendulum.or \$ L = \dfrac{gT^2}{4 \pi^2}\$The time period of the simple pendulum which ticks seconds is \$2\$s.\$\therefore T = 2s\$Substituting in (i), we get\$L = \dfrac{(9.8 m s^{-2})(2s)^2}{4 \times (3.14)^2} = 1m\$

Q: The length of a second pendulum at the surface of earth is \$1\ m\$. The length of second pendulum at the surface of moon, where \$g\$ is \$\dfrac{1}{6} th\$ that of earth's surface.

We know that the time period of a second pendulum is \$2s\$ .At earth , \$T=2\pi \sqrt{l_{e}/g_{e}}\$ \$2=2\pi\sqrt{l_{e}/g_{e}}\$or \$g_{e}=\pi^{2}l_{e}\$ ...............................eq1At moon , \$T=2\pi \sqrt{l_{m}/g_{m}}\$or \$2=2\pi \sqrt{l_{m}/(g_{e}/6)}\$ , given \$g_{m}=g_{e}/6\$or \$2=2\pi\sqrt{6l_{m}/\pi^{2}l_{e}}\$ , putting the value of \$g_{m}\$ from eq1or \$l_{m}=l_{e}/6\$Now , given \$l_{e}=1m\$Hence , \$l_{m}=1/6m\$

Q: If effective length of a simple pendulum is equal to radius of earth (R), time period will be

formula,\$T=2\pi \sqrt{\dfrac{LR}{g(R+L)}}\$If the effective length of a simple pendulum is equal to the radius of the earth (R),then,\$T=2\pi \sqrt{\dfrac{R\times R}{g(R+R)}}\$\$=2\pi \sqrt{\dfrac{R^2}{g(2R)}}\$\$\therefore T=2\pi \sqrt{\dfrac{R}{2g}}\$

Q: A girl is swinging in a swing in a sitting position. How will the period of the swing be affected if she stands up?

Let the swinging be a simple pendulum,So the time period of the simple pendulum is:\$T=2\pi\sqrt{\dfrac{L}{g}}\$Where L is the length,Since length of the simple pendulum is equal to the distance from the point of suspension to CG of the oscellting body,When the girl stands up the distance of CG from the point of suspension will decrease.Therefore time period of the oscillation also decreases.

Q: The length of second pendulum is :

Time period of second pendulum \$T = 2 \ s\$Time period \$T =2\pi \sqrt{\dfrac{l}{g}}\$where \$l\$ is the length of pendulum.\$\therefore\$ \$2 = 2\pi\sqrt{\dfrac{l}{9.8}}\$\$\implies\ l = 0.994 \ m = 99.4 \ cm\$

Q: A simple pendulum is being used to determine the value of gravitational acceleration \$g\$ at a certain place. The length of the pendulum is \$25.0cm\$ and a stopwatch with \$1s\$ resolution measures the time taken for \$40\$ oscillation to be \$50s\$. The accuracy in \$g\$ is:

For simple pendulum\$T = 2\pi \sqrt{\dfrac{l}{g}}\$so, \$\dfrac{\Delta T}{T} = \dfrac{1}{2} \left(\dfrac{\Delta l}{l} + \dfrac{\Delta g}{g}\right)\$Givn \$\Delta T = 1 \ sec\$ \$T = 50 sec\$\$\Delta l = 0.1 cm\$\$l = 25 cm\$\$\dfrac{\Delta g}{g} = \dfrac{2\Delta T}{T} + \dfrac{\Delta l}{l}\$\$\dfrac{\Delta g}{g} = 2\times \dfrac{1}{50} + \dfrac{0.1}{25} = \dfrac{1.1}{25}\$\$\dfrac{\Delta g}{g} = \dfrac{1.1}{25}\$\$\% \dfrac{\Delta g}{g} = \dfrac{1.1}{25} \times 100 = 4.4 \%\$

Q: Four pendulums A, B, C and D are suspended from the same elastic support as shown in figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement.

Since length of pendulum A and C is same and \$T = 2 \pi \sqrt{\frac{L}{g}}\$, hence their time period is same and they will have same frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude.

Q: If length of pendulum is increased by 2%. The time period will

Time period of pendulum is given by \$T = 2\pi\sqrt{\dfrac{l}{g}}\$Error in time period \$\dfrac{\Delta T}{T}\times 100 = \dfrac{1}{2}\times \$ \$\dfrac{\Delta l}{l}\times 100\$Given : \$\dfrac{\Delta l}{l}\times 100 = 2\$%\$\implies \ \$ \$\dfrac{\Delta T}{T}\times 100 = \dfrac{1}{2}\times \$ \$2 = 1\$%Hence the time period increases by 1% if the length of pendulum is increased by 2%.

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