Constants and other possibly useful numbers g = 9.8m / (s ^ 2) e = 1.6 * 10 ^ – 19 * C R = k_{B}*N_{A} = 8.31 * J/(molK) G = 6.67 * 10 ^ – 11 * (N * m ^ 2)/(k * g ^ 2) ħ = h/(2π) = 1.05 × 10-34 JS 0K = – 273 deg * C c = 3 * 10 ^ 8 * m / s N_{A} = 6.02 * 10 ^ 23 / m * ol 1atm = 1.01 * 10 ^ 5 * Pa m_{e} = 9.1 * 10 ^ – 31 * kg k_{B} = 1.38 * 10 ^ – 23 * J / K 1eV = 1.6 * 10 ^ – 19 * J Specific Heat (c) J/(kg·K) kcal/(kg°C) Aluminum 900 0.215 Constants Copper 387 0.0924 Gold 129 0.0308 Iron, steel 452 0.108 1.000 Water 4186 1. Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63m / (s ^ 2) .
The Correct Answer and Explanation is :
The period TT of a simple pendulum is given by the formula:
T=2πLgT = 2\pi \sqrt{\frac{L}{g}}
where:
- TT is the period,
- LL is the length of the pendulum,
- gg is the acceleration due to gravity.
This equation shows that the period is directly proportional to the square root of the pendulum’s length and inversely proportional to the square root of the acceleration due to gravity.
Ratio of Periods:
When a pendulum is moved from Earth to the Moon, the acceleration due to gravity changes from Earth’s gEarth=9.8 m/s2g_{\text{Earth}} = 9.8 \, \text{m/s}^2 to the Moon’s gMoon=1.63 m/s2g_{\text{Moon}} = 1.63 \, \text{m/s}^2.
The ratio of the new period (TMoonT_{\text{Moon}}) to the old period (TEarthT_{\text{Earth}}) is:
\frac{T_{\text{Moon}}}{T_{\text{Earth}}} = \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Moon}}}
Substituting the values:
TMoonTEarth=9.81.63≈6.01\frac{T_{\text{Moon}}}{T_{\text{Earth}}} = \sqrt{\frac{9.8}{1.63} \approx 6.01}
TMoonTEarth≈2.45\frac{T_{\text{Moon}}}{T_{\text{Earth}}} \approx 2.45
Therefore, the period of the pendulum on the Moon is approximately 2.45 times longer than on Earth.
Explanation:
The period of a pendulum depends on the acceleration due to gravity. On Earth, with gEarth=9.8 m/s2g_{\text{Earth}} = 9.8 \, \text{m/s}^2, the period is shorter because the gravitational pull is stronger, causing the pendulum to swing back and forth more quickly. On the Moon, with gMoon=1.63 m/s2g_{\text{Moon}} = 1.63 \, \text{m/s}^2, the weaker gravitational pull results in a longer period, as the pendulum swings more slowly.
This relationship highlights how gravitational acceleration influences the motion of pendulums, and by extension, the measurement of time.