Question

In the figure given below and idle pendulum oscillates in vacuum between two extreme position A and B if it goes from position 82 position B and then return from position B2 mean position see in 3 seconds then it will complete 3.5 oxidation and how many seconds

Answer 1

Let's call the period of the pendulum TT. The pendulum moves from point A to point B in half of a period 12T12T. It then moves from point B to C in half of half of TT, which is 12⋅12T12⋅12T or 14T14T. So it moves from point A to B then to C in 12T+1412T+14 which is equivalent to 34T34T.

The question says all this is done in 33seconds. This means

34T=334T=3

T=4T=4  seconds seconds

So, the period was 4 seconds4 seconds. Period itself is defined as the ratio of the time taken tt to the number of oscillations nn in that time.

T=tnT=tn

Since the number of oscillations nn is 3.5 cycles3.5 cycles, and the period is 4 seconds4 seconds, inputting these values;

4=t3.54=t3.5

t=4×3.5=14 secondst=4×3.5=14 seconds

To convert this to minutes, divide by 6060.

t=14/60=0.23 minutes


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