The time period of a pendulum is measured to be 3.0 seconds in the inertial reference frame of the pendulum. What is its period measured by an observer moving at a speed of 0.95c with respect to the pendulum?
Correct answer: D. 9.6 s
- A. 1.2 s
- B. 3.4 s
- C. 8.1 s
- D. 9.6 s
Explanation
The correct option is D. 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 .- Using the formula- The time dilation formula is given by,- T =T0 /√1−(v2/c2)where,- T is the time observed- T0 is the time observed at rest- v is the velocity of the object- c is the velocity of light in a vacuum (3 × 108 m/s2)T= 3/√1−[(0.95c)2/c2] - T= 3/√1−(0.90c2/c2)- T= 3/√1-0.90- T= 3/√0.10- Take 0.10 to 0.9 to solve easily- T= 3/√0.9- T= 3/0.3- T= 10 seconds 9.6 seconds
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About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
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