Question

Найти период и частоту колебаний матем маятника длина которого равна 0,634м

Answer 1

T=2*pi*sqrt(L/g)=6,28*(0,634/9,8)=1,6 c
v=1/T=0,625 Гц

Answer 2

Period and Frequency of Oscillations of a Pendulum

To find the period and frequency of oscillations of a pendulum with a length of 0.634m, we can use the following formulas:

1. Period (T): The time taken for one complete oscillation. - The formula for the period of a pendulum is given by: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where: - \( T \) = period of the pendulum - \( \pi \) = mathematical constant (approximately 3.14159) - \( L \) = length of the pendulum - \( g \) = acceleration due to gravity (approximately 9.81 m/s\(^2\))

2. Frequency (f): The number of oscillations per unit time. - The formula for the frequency of a pendulum is given by: \[ f = \frac{1}{T} \] where: - \( f \) = frequency of the pendulum - \( T \) = period of the pendulum

Calculations

Using the given length of the pendulum (0.634m) and the standard acceleration due to gravity (9.81 m/s\(^2\)), we can calculate the period and frequency of oscillations.

1. Period (T): \[ T = 2\pi \sqrt{\frac{0.634}{9.81}} \] \[ T \approx 2\pi \times \sqrt{0.0647} \] \[ T \approx 2\pi \times 0.2545 \] \[ T \approx 1.599 \, \text{s} \]

2. Frequency (f): \[ f = \frac{1}{1.599} \] \[ f \approx 0.625 \, \text{Hz} \]

So, for a pendulum with a length of 0.634m, the period of oscillation is approximately 1.599 seconds, and the frequency of oscillation is approximately 0.625 Hertz.

These calculations are based on the formulas for the period and frequency of a pendulum, and the values of \( \pi \) and \( g \).