Physics, asked by 0Adidas, 1 year ago

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 ?

Answers

Answered by rohit710
73
Heya.......!!!!

_____________________

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
 \gamma

We know that proper time ( ∆t ) is

=>
 \gamma \times d(t _{p}) \\ \\ substituting \: the \: values \: \\ \\ also \: \gamma \: = \: \frac{1}{ \sqrt{1 - \frac{v {}^{2} }{c {}^{2} } } } \\ \\ now \: putting \: the \: vaues \: \\ \\ = > \: \: \frac{1}{ \sqrt{ 1 - \frac{(0.95c) {}^{2} }{(3 \times 10 {}^{8} ) {}^{2} } } } \\ \\ its \: a \: long \: calculation \: \\ \\ = > \: \frac{1}{ \sqrt{1 - 0.902} } \times \: t _{p} \\ = > \: 3.2 \times 3 \\ \\ = > 9.6 \: sec

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 ☺

0Adidas: Thank you very much !
rohit710: Pls don't say thanks
rohit710: :))
rohit710: Note that d(t) = ∆t
rohit710: it is 0.95c^2. in calculation
Answered by BrainlyFlash156
7

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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

 \gamma

We know that proper time ( ∆t ) is

=>

 \gamma \times d(t _{p}) \\ \\ substituting \: the \: values \: \\ \\ also \: \gamma \: = \: \frac{1}{ \sqrt{1 - \frac{v {}^{2} }{c {}^{2} } } } \\ \\ now \: putting \: the \: vaues \: \\ \\ = > \: \: \frac{1}{ \sqrt{ 1 - \frac{(0.95c) {}^{2} }{(3 \times 10 {}^{8} ) {}^{2} } } } \\ \\ its \: a \: long \: calculation \: \\ \\ = > \: \frac{1}{ \sqrt{1 - 0.902} } \times \: t _{p} \\ = > \: 3.2 \times 3 \\ \\ = > 9.6 \: sec

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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