Solved Numerical Problem No.2 ( Unit-18: Dawn of Modern Physics); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

The time period of a pendulum is measured to be 3s in the Inertial frame of the pendulum. What is the period when measured by an observer moving with a speed of 0.95c with respect to the pendulum? (Answer: 9.65)

Given:

Proper time period = t₀ = 3 s

Speed of observer = v = 0.95 c 

To Find:

Measure Time Period = t =?

Solution:

The time dilation formula in special theory of relativity is given by: 

t = `\frac {t₀}{sqrt {1 - frac {v^2}{c^2}}}`

by putting values

t = `\frac {3 s}{sqrt {1 - frac {(0.95c)^2}{c^2}}}`

c² will be canceled so,

t = `\frac {3 s}{sqrt {1 - 0.9025}}`

t = `\frac {3 s}{sqrt {0.0875}}`

t = `\frac {3 s}{0.312}`

t = 9.615 s    -----------------Ans.

Thus, at the speed of observer = 0.95 c (velocity of light), time dilation in the period of the pendulum will occur equal to 9.615 s 

Similar Questions:

A particle called the pion lives on the average only about 2.6 ｘ10⁻⁸ s when at rest in the laboratory. It then changes to another form. How long would such a particle live when shooting through the space at 0.95c?[Ans. 8.3 x 10⁻⁸ s)

The length of a space ship is measured to be exactly one-third of its proper length. What is the speed of the space ship relative to the observer? (Answer: 0.9428 c)

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