Problem

The time period of a pendulum is measured to be 3 seconds in the inertial frame of the pendulum. What is the time period when measured by an observer moving with a speed of $0.95c$ with respect to the pendulum?

---

Data

• Proper time: $\Delta t_p = 3.00 \, \mathrm{s}$
• Speed of the observer relative to the pendulum: $v = 0.95c$
• Speed of light: $c$

To Find: Time dilation effect, dilated time ($\Delta t$).

---

Prerequisite Concepts

The relativistic time dilation equation is:

$$\Delta t = \frac{\Delta t_p}{\sqrt{1 - \frac{v^2}{c^2}}}$$

where:

• $\Delta t$ is the dilated time observed,
• $\Delta t_p$ is the proper time (time measured in the pendulum’s rest frame),
• $v$ is the relative speed between the observer and the pendulum,
• $c$ is the speed of light.

---

Solution

1. Substitute the known values into the time dilation formula:

$$\Delta t = \frac{\Delta t_p}{\sqrt{1 - \frac{v^2}{c^2}}}$$ $$\Delta t = \frac{3.00 \, \mathrm{s}}{\sqrt{1 - \frac{(0.95c)^2}{c^2}}}.$$

2. Simplify the denominator:

$$\Delta t = \frac{3.00 \, \mathrm{s}}{\sqrt{1 - 0.9025}}.$$

3. Calculate the square root:

$$\Delta t = \frac{3.00 \, \mathrm{s}}{\sqrt{0.0975}}.$$

4. Simplify further:

$$\Delta t = \frac{3.00 \, \mathrm{s}}{0.3123}.$$

5. Perform the division:

$$\Delta t = 9.61 \, \mathrm{s}.$$

---

Answer

The time period measured by the moving observer is approximately $\Delta t = 9.61 \, \mathrm{s}$.