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.\\)

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}}\\)

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\\)

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\\)

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}}\\)

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.

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\\)

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 \%\\)

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.

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%.

Simple Pendulum | Definition, Examples, Diagrams

definition

Working of a simple pendulum

A simple pendulum has a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string. When the bob from its mean position is pulled to one side and then released, the pendulum is set to motion and the bob moves alternately on either side of its mean position.

definition

Terms related to Simple Pendulum

These are important terms pertaining to simple pendulum:
1. Oscillation: One complete to and fro motion of the pendulum is called one oscillation
2. Time period: This is the time taken to complete one oscillation.
3. Frequency of oscillation: It is the number of oscillations made in one second.

definition

Effective Length of Pendulum

It is the distance of point of oscillation (i.e. the centre of gravity of the bob) from the point of suspension.

definition

Working of Simple Pendulum

Construction : A simple pendulum consists of a small metallic ball or a piece of stone suspended from a rigid stand by a thread [Fig(a)]. The metallic ball is attached to the pendulum called bob.
Working :Fig(a) shows the pendulum at rest in its mean position. When the bob of the pendulum is released after taking it slightly to one side, it begins to move to and fro [Fig (b)]. The to and fro motion of a simple pendulum is an example of a periodic or an oscillatory motion.
The pendulum is said to have completed one oscillation when its bob,starting from its mean position O, moves to A, to B and back to O.
The pendulum also completes one oscillation when its bob moves from one extreme position A to the other extreme position B and comes back to A.

definition

Define Bob

Bob : A simple pendulum consists of a small metallic ball or a piece of stone suspended from a rigid stand by a thread called the bob of the pendulum.
Time period of pendulum doesn't depend on the size or mass of the pendulum, it depends on the length of the pendulum and factor of earth's gravity.

Simple Pendulum

definition

Working of a simple pendulum

definition

Terms related to Simple Pendulum

definition

Effective Length of Pendulum

definition

Working of Simple Pendulum

definition

Define Bob

Quick Summary With Stories

Simple Pendulum

Working of a simple pendulum

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 $\frac{T _{2}}{T _{1}}$ is:

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.

The length of the simple pendulum which ticks seconds is:

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 $\frac{1}{6} t h$ that of earth's surface.

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

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

The length of second pendulum is :

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

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.

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

definition

Working of a simple pendulum

definition

Terms related to Simple Pendulum

definition

Effective Length of Pendulum

definition

Working of Simple Pendulum

definition

Define Bob

Quick Summary With Stories

Simple Pendulum

Working of a simple pendulum

A simple pendulum has a time period T1​ when on the earth's surface and T2​ when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T1​T2​​ is:

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.

The length of the simple pendulum which ticks seconds is:

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 61​th that of earth's surface.

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

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

The length of second pendulum is :

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:

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.

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

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 $\frac{T _{2}}{T _{1}}$ is:

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 $\frac{1}{6} t h$ that of earth's surface.

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