The Period of a Pendulum is measured to be 3.0 sec in the reference frame of the pendulum What is the period when measured by an observer moving at speed of 0.95 c relative to the pendulum ?
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Heya.......!!!!
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This is an Easy question of Chapter 'Relativity '
= It is given that the observer is moving at a speed of 0.95 c . So let us assume that instead of observer ,, the Pendulum is moving at 0.95 c and the observer is at rest .
In the question ' Proper time ' is given
=> ∆t = 3 sec
Now , moving clock runs slowly than stationary clock by a factor of

We know that proper time ( ∆t ) is
=>

Final answer => 9.6 sec
9.6 sec is the period when measured by an observer moving at a speed 0.95 c relative to pendulum .
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Hope It Helps You ☺
_____________________
This is an Easy question of Chapter 'Relativity '
= It is given that the observer is moving at a speed of 0.95 c . So let us assume that instead of observer ,, the Pendulum is moving at 0.95 c and the observer is at rest .
In the question ' Proper time ' is given
=> ∆t = 3 sec
Now , moving clock runs slowly than stationary clock by a factor of
We know that proper time ( ∆t ) is
=>
Final answer => 9.6 sec
9.6 sec is the period when measured by an observer moving at a speed 0.95 c relative to pendulum .
========================
Hope It Helps You ☺
0Adidas:
Thank you very much !
Answered by
7
This is an Easy question of Chapter 'Relativity '
= It is given that the observer is moving at a speed of 0.95 c . So let us assume that instead of observer ,, the Pendulum is moving at 0.95 c and the observer is at rest .
In the question ' Proper time ' is given
=> ∆t = 3 sec
Now , moving clock runs slowly than stationary clock by a factor of
We know that proper time ( ∆t ) is
=>
Final answer => 9.6 sec
9.6 sec is the period when measured by an observer moving at a speed 0.95 c relative to pendulum .
HOPE SO IT WILL HELP.....
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