The time period of a simple pendulum is 2 s. What will be its length on the Moon if gm = ge/6, where ge = 10 m/s²? A pendulum of length 0.99 m is taken to the Moon by an astronaut.... The time period of a simple pendulum is 2 s. What will be its length on the Moon if gm = ge/6, where ge = 10 m/s²? A pendulum of length 0.99 m is taken to the Moon by an astronaut. The period of the pendulum is 4.9 s. What is the value of g on the surface of the Moon? Read more

Steps to Solve

1. Formula for the Period of a Pendulum
   The time period $T$ of a simple pendulum is given by the formula:
   $$ T = 2\pi \sqrt{\frac{L}{g}} $$
   where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

2. Finding Length on Earth
   Using the period of the pendulum on Earth ($T = 2 , s$), we can rearrange the formula to solve for $L$:
   $$ L = \frac{g T^2}{4\pi^2} $$
   We know $g_e = 10 , m/s^2$, so substituting the values:
   $$ L = \frac{10 \times (2)^2}{4\pi^2} $$

3. Calculate Length on Earth
   Calculating the length:
   $$ L = \frac{10 \times 4}{4\pi^2} = \frac{10}{\pi^2} \approx 1.02 , m $$

4. Finding Length on Moon
   The length of the pendulum on the Moon when $g_m = \frac{g_e}{6}$ is given by: $$ g_m = \frac{10}{6} , m/s^2 \approx 1.67 , m/s^2 $$ Now substituting into the formula for the Moon: $$ L_m = \frac{g_m T_m^2}{4\pi^2} $$

5. Substituting Time Period on Moon
   Given the time period $T_m = 4.9 , s$: $$ L_m = \frac{1.67 \times (4.9)^2}{4\pi^2} $$

6. Calculate Length on Moon
   Carrying out the calculation: $$ L_m \approx \frac{1.67 \times 24.01}{4\pi^2} \approx \frac{40.0967}{39.4784} \approx 1.016 , m $$

7. Finding Value of g on Moon
   Now we rearrange the pendulum formula to find $g$ given $L = 0.99 , m$ and $T = 4.9 , s$:
   $$ g = \frac{4\pi^2 L}{T^2} $$
   Substituting the values: $$ g = \frac{4\pi^2 \times 0.99}{(4.9)^2} $$

8. Calculate g on Moon
   Calculating: $$ g \approx \frac{4 \times 9.8696 \times 0.99}{24.01} \approx \frac{39.0964}{24.01} \approx 1.63 , m/s^2 $$