# Torque in oscillation

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162 views In this , I could not understand wny in oscillation both the torque would be equal as given in the solution .

asked Mar 5, 2017
It is not obvious that torque should be the same. However, the PE should be the same, because KE is zero at both extreme points, and energy is conserved. (I assume it is a condition that the string being vertical is also an extreme point, otherwise I do not think this is necessary. If $qE \gg mg$ then the string will oscillate about the $E$ direction, whereas if $qE \ll mg$ it will oscillate about the $g$ direction.)
I could not understand why they will oscillate and in what direction .
Perhaps they do not oscillate. Perhaps this is a static problem, then it is obvious that the torques must be equal and opposite. But in that case the string cannot be parallel to the electric field, as the question states. Another poorly worded question, and confusing solution.

If the pendulum does oscillate, it does not have to stop when the string is parallel to the electric and gravitational fields. This seems to be a condition in the question, or an assumption in the answer.

## 1 Answer

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it is not clear if the question is asking about the static equilibrium of the pendulum or about its oscillations. In either case we can find the equilibrium angle $\alpha$ of the string to the vertical from a vector diagram. The accelerations due to gravity (g) and due to the constant electric field (qE/m) add as vectors to give a new effective acceleration due to gravity (g').

The pendulum will hang along the direction of g', or oscillate about the line of g', at equal angles to it left and right. g' is only the bisector of g and qE/m if g=qE/m. Then $\alpha=\frac12 \theta$. So the amplitude of the swing will only coincide with the directions g and qE/m if they are at equal angles from g. This requires g=qE/m.

answered Mar 7, 2017 by (28,746 points)
edited Mar 7, 2017
So there is no relation with torque
The torque calculation is valid, but it assumes that the extremes of oscillation lie along g and E.

The question seems to say that the pendulum string rises up to the direction of E. That is impossible statically ; it can only happen in the limit of qE/mg -->infinity. Another interpretation is that the string reaches the direction of E dynamically, at the end of its swing. However, the question says nothing about the other extreme of the swing. In my diagram it will swing past the direction of g, making the same angles left and right of g'.

The solution assumes that the 2nd extreme of swing is along g, although there is nothing in the question to justify this assumption.

Torque is equal (and opposite) at both ends of the swing for any pendulum. So the calculation in the given solution is valid.

The torque calculation could be used for the static case also - ie the equilibrium position when the oscillations die out. Then the string lies along g' (left diagram). Torque due to g is $mgl \sin\alpha$ clockwise, torque due to E is $qEl \sin(\theta-\alpha)$ anticlockwise. These must be equal for any static equilibrium situation, not only the special situation in this question, in which value of E allows the pendulum to swing between the directions of g and E.
Thanks , I got it