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A Simple Pendulum of Length L is Suspended Through the Ceiling of an Elevator. Find the Time Period of Small Oscillations - Physics

Sum

A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with and acceleration a0(b) is going down with an acceleration a0 and (c) is moving with a uniform velocity.

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Solution

The length of the simple pendulum is l.
‚ÄčLet x be the displacement of the simple pendulum..
(a)

From the diagram, the driving forces f is given by,
f = m(g + a0)sinθ                 ...(1)
Acceleration (a) of the elevator is given by,

\[a = \frac{f}{m}\] 

\[   = \left( g + a_0 \right)\sin\theta\] 

\[   = \left( g + a_0 \right)\frac{x}{l}       \left( \text{From  the  diagram } \sin\theta = \frac{x}{l} \right)_{}\]

[ when θ is very small, sin θ → θ = x/l]

\[\therefore a = \left( \frac{g + a_0}{l} \right)x\]  ...(2)

As the acceleration is directly proportional to displacement, the pendulum executes S.H.M.
Comparing equation (2) with the expression a =\[\omega^2 x\],we get:

\[\omega^2  = \frac{g + a_0}{l}\]

Thus, time period of small oscillations when elevator is going upward(T) will be:

\[T = 2\pi\sqrt{\frac{l}{g + a_0}}\]
When the elevator moves downwards with acceleration a0,
      Driving force (F) is given by,
      F = m(g − a0)sinθ
On comparing the above equation with the expression, F = ma,\[\text{Acceleration},   a = \left( g - a_0 \right)  sin\theta = \frac{\left( g - a_0 \right)x}{l} =  \omega^2 x\] \[\text{Time  period  of  elevator  when  it  is  moving  downward}\left( T' \right) \text{ is  given  by,} \] 

\[T' = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{l}{g - a_0}}\]

(c) When the elevator moves with uniform velocity, i.e. a0 = 0,
For a simple pendulum, the driving force \[\left( F \right)\] is given by,

\[F = \frac{mgx}{l}\] 

\[\text{Comparing  the  above  equation  with  the  expression,   F = ma,   we  get: }\] \[a = \frac{gx}{l}\] \[ \Rightarrow \frac{x}{a} = \frac{l}{g}\] \[T = 2\pi\sqrt{\frac{\text{displacement}}{\text{Acceleration}}}\] \[   = 2\pi\sqrt{\frac{l}{g}}\]

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 12 Simple Harmonics Motion
Q 43 | Page 255
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