Answer
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Hint: The period of a pendulum is directly proportional to the length of the pendulum and inversely proportional to acceleration due to gravity. Use the relation between $g$ and $E$ to derive the period of the pendulum.
Complete step by step solution:
The period of a pendulum is defined as the time taken by the pendulum to complete one oscillation. Period of a pendulum is independent of factors like mass of the sphere, temperature etc. It depends on the factors like acceleration due to gravity and length of the pendulum. It is mathematically expressed as:
$T = 2\pi \sqrt {\dfrac{l}{g}} $
where, $l$ = length of the pendulum.
$g$ = acceleration due to gravity.
Motion of simple pendulum is simple harmonic motion and the angular frequency of the motion is given by:
$\omega = \sqrt {\dfrac{g}{l}} $
where $l$ = length of the pendulum.
$g$ = acceleration due to gravity.
It is given that the pendulum is under the effect of uniform field $E$, to calculate the effect of electric field on period of pendulum we have to relate the net acceleration obtained due to the uniform field.
Let the charge developed on the pendulum be $q$. The total net force acting on the pendulum is: ${F_{net}} = qE + mg$
This gives, ${F_{net}} = qE + 0$ since the effect of gravitational force is neglected
$ \Rightarrow {F_{net}} = qE$
Also, $F = ma$
Now relating the above two equations, we get $ma = qE$
Here, $a = $ acceleration $ = g$ Therefore, $mg = qE$
$ \Rightarrow g = \dfrac{{qE}}{m}$
Now putting the value of $g$ in the pendulum formula we get:
$T = 2\pi \sqrt {\dfrac{l}{{\dfrac{{qE}}{m}}}} $
$ \Rightarrow T = 2\pi \sqrt {\dfrac{{lm}}{{qE}}} $
Therefore, the required value of period of the pendulum is: $T = 2\pi \sqrt {\dfrac{{lm}}{{qE}}}.$
Note: Period of the pendulum is obtained from the calculated angular frequency from the required expression. Pendulums are used as clocks and can be seen in swings. It follows the simple harmonic motion i.e. the motion repeats itself after a certain interval. A simple pendulum consists of a simple bob of negligible mass attached to a string.
Complete step by step solution:
The period of a pendulum is defined as the time taken by the pendulum to complete one oscillation. Period of a pendulum is independent of factors like mass of the sphere, temperature etc. It depends on the factors like acceleration due to gravity and length of the pendulum. It is mathematically expressed as:
$T = 2\pi \sqrt {\dfrac{l}{g}} $
where, $l$ = length of the pendulum.
$g$ = acceleration due to gravity.
Motion of simple pendulum is simple harmonic motion and the angular frequency of the motion is given by:
$\omega = \sqrt {\dfrac{g}{l}} $
where $l$ = length of the pendulum.
$g$ = acceleration due to gravity.
It is given that the pendulum is under the effect of uniform field $E$, to calculate the effect of electric field on period of pendulum we have to relate the net acceleration obtained due to the uniform field.
Let the charge developed on the pendulum be $q$. The total net force acting on the pendulum is: ${F_{net}} = qE + mg$
This gives, ${F_{net}} = qE + 0$ since the effect of gravitational force is neglected
$ \Rightarrow {F_{net}} = qE$
Also, $F = ma$
Now relating the above two equations, we get $ma = qE$
Here, $a = $ acceleration $ = g$ Therefore, $mg = qE$
$ \Rightarrow g = \dfrac{{qE}}{m}$
Now putting the value of $g$ in the pendulum formula we get:
$T = 2\pi \sqrt {\dfrac{l}{{\dfrac{{qE}}{m}}}} $
$ \Rightarrow T = 2\pi \sqrt {\dfrac{{lm}}{{qE}}} $
Therefore, the required value of period of the pendulum is: $T = 2\pi \sqrt {\dfrac{{lm}}{{qE}}}.$
Note: Period of the pendulum is obtained from the calculated angular frequency from the required expression. Pendulums are used as clocks and can be seen in swings. It follows the simple harmonic motion i.e. the motion repeats itself after a certain interval. A simple pendulum consists of a simple bob of negligible mass attached to a string.
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