The period of an oscillating weight on a spring in an inertial frame of reference is 2.5 s.What would be its speed if it were to move by an observer who measures its period as 3.4 s? (c = 3.00 × 10⁸ m/s) A)1.5 × l0⁸ m/s B)2.0 × 10⁸ m/s C)2.8 × l0⁸ m/s D)6.8 × 10⁷ m/s E)1.3 × 10⁸ m/s

Question

Quizplus · Accepted Answer

The speed of the observer relative to the inertial frame can be found using the relativistic Doppler effect formula for time dilation: $v = c \sqrt{1 - \left(\frac{T_0}{T}\right)^2}$, where $T_0$ is the period in the inertial frame (2.5 s), $T$ is the observed period (3.4 s), and $c$ is the speed of light. Plugging in the values, we find that the speed that matches the given conditions is $2.0 \times 10^8 m/s$, which corresponds to choice B.

The Period of an Oscillating Weight on a Spring in an Inertial