i just started studying special theory of relativity and am confused a lot. its first consequence,that is moving clocks run slow confuses me a lot. if two events happen then the time difference between them as measured by a clock in a frame moving at .9c would be lesser than the time measured by a similar clock which is in a frame that is at rest.(adsbygoogle = window.adsbygoogle || []).push({});

now should i first specify here that which is the proper time? the proper time,as i understand it, would be the time measured by the clock in whose frame the two events occured at the same position. right?

lets say that the events were stationery with respect to the frame which i see to be moving at .9c. (i am at rest.) now since i am at rest and we know that moving clocks run slow then i will say that the time measured in the frame of refrence of the events is less than the actual time(as measured by the clock in my frame.) right? so the moving clock measured 5s but mine measured 10s.

but then a thought occured to me that to someone who is in the moving frame(that is who is at rest in the 0.9c frame) i will look like moving with -.9c. and he has equal right in saying that my clock is slow(the one who seems to be moving at -.9c) and his is normal. so if his measures 5s then mine measures less that is 2.5s.

now one of the situations is obviously wrong. but i cant understand which one? kindly explain how to tell proper time from "improper" time and which situation is correct?

actually this doubt arose because of a numerical in one book i was reading. here is it:

Question :

The period of a pendulum is measured to be 3.00 s in the reference frame of the pendulum. What is the period when

measured by an observer moving at a speed of 0.960c relative to the pendulum?

SOLUTION :

Concept: Let’s change frames of reference. Instead of the observer moving at 0.960c, we can take the equivalent point of view that the observer is at rest and the pendulum is moving at 0.960c past the stationary observer. Hence, the pendulum is an example of a clock moving at high speed with respect to an observer.

Analyze : The proper time interval, measured in the rest frame of the pendulum, is Δtp = 3.00 s.

Use time-dilation equation to find the dilated time interval: Δt = Δtp*omega

omega = 1/((1-(v^2)/(c^2))^0.5)

the time interval after calculation comes to be 9.57s.

now i say that instead of the observer which was moving, his time is slow and his clock must measure less than 3.00s but the book says otherwise.

where do i go wrong? do explain.

thanks in advance

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# Time dilation in the period of a pendulum

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