Question

Как изменится частота малых колебаний математического маятника если длину его нити уменьшить в 4 раза а массу груза увеличить в 5 раз?

Answer 1

Дано L2=L/4     m2=5*m     v2- ?

v2=1/T2=1/2*π*√L/g=1/2*π*√L/4*g=2*v1

частота увеличится в 2 раза ( от массы частота не  зависит)

Answer 2

Calculating the Change in Frequency of a Pendulum

When the length of the pendulum is reduced by a factor of 4 and the mass of the bob is increased by a factor of 5, the frequency of small oscillations of the pendulum will change. Let's calculate this change step by step.

1. Original Frequency of the Pendulum: The frequency of small oscillations of a pendulum is given by the formula: \[ f = \frac{1}{2\pi}\sqrt{\frac{g}{L}} \] where: - \( f \) = frequency of the pendulum - \( g \) = acceleration due to gravity - \( L \) = length of the pendulum

Let's assume the original length of the pendulum is \( L_0 \) and the original mass of the bob is \( m_0 \).

2. New Length and Mass: After the changes, the length of the pendulum becomes \( \frac{L_0}{4} \) and the mass of the bob becomes \( 5m_0 \).

3. Calculating the New Frequency: Using the formula for the frequency of a pendulum, we can calculate the new frequency after the changes.

The new frequency, \( f' \), is given by: \[ f' = \frac{1}{2\pi}\sqrt{\frac{g}{\frac{L_0}{4}}} \]

Substituting the new values, we get: \[ f' = \frac{1}{2\pi}\sqrt{4\frac{g}{L_0}} = 2\frac{1}{2\pi}\sqrt{\frac{g}{L_0}} \]

This means that the new frequency is 2 times the original frequency.

Conclusion

Therefore, when the length of the pendulum is reduced by a factor of 4 and the mass of the bob is increased by a factor of 5, the frequency of small oscillations of the pendulum will increase by a factor of 2. This conclusion is based on the calculations using the formula for the frequency of a pendulum and the given changes in length and mass of the bob.