Understanding Pendulum Oscillation and Constant Phase Difference

When two pendulums oscillate, we can describe their motion using concepts like time period, frequency, and phase. The time period (\(T\)) is the time taken for one complete back-and-forth swing. The frequency (\(f\)) is the number of oscillations per second, and it's the reciprocal of the time period, \(f = \frac{1}{T}\). Phase refers to the position and direction of motion of the pendulum at any given time relative to its starting point or another oscillating object.

A constant phase difference between two oscillations means that the difference in their phase angles remains the same over time. If two pendulums oscillate with a constant phase difference, it indicates a specific relationship between their motions that doesn't change as time progresses.

Constant Phase Difference and Time Period

For two oscillations to maintain a constant phase difference, their frequencies must be exactly the same. Think about it: if the frequencies were different, one pendulum would be oscillating faster or slower than the other. This difference in speed of oscillation would cause the phase difference between them to continuously change, not remain constant.

Let the phase of the first pendulum be \(\phi_1(t) = \omega_1 t + \delta_1\) and the phase of the second pendulum be \(\phi_2(t) = \omega_2 t + \delta_2\), where \(\omega_1\) and \(\omega_2\) are their angular frequencies and \(\delta_1\) and \(\delta_2\) are initial phases. The phase difference is \(\Delta \phi(t) = \phi_2(t) - \phi_1(t) = (\omega_2 - \omega_1)t + (\delta_2 - \delta_1)\).

For \(\Delta \phi(t)\) to be constant, the term that depends on time, \((\omega_2 - \omega_1)t\), must be zero for all \(t\). This can only happen if \(\omega_2 - \omega_1 = 0\), which means \(\omega_1 = \omega_2\).

Angular frequency (\(\omega\)) is related to frequency (\(f\)) by \(\omega = 2\pi f\). So, if \(\omega_1 = \omega_2\), then \(2\pi f_1 = 2\pi f_2\), which simplifies to \(f_1 = f_2\). The frequency is also related to the time period by \(f = \frac{1}{T}\). Therefore, \(f_1 = f_2\) implies \(\frac{1}{T_1} = \frac{1}{T_2}\), which means \(T_1 = T_2\).

In summary, a constant phase difference between two pendulums during oscillation is only possible if they have the same frequency, and consequently, the same time period.

Calculating the Other Pendulum's Time Period

The question states that two pendulums oscillate with a constant phase difference of 90°. This constant phase difference tells us that their frequencies are identical. One pendulum has a time period of 2 seconds.

• Time period of the first pendulum, \(T_1 = 2\) sec.
• The phase difference between the two pendulums is constant.
• Because the phase difference is constant, the frequency of the first pendulum (\(f_1\)) must be equal to the frequency of the second pendulum (\(f_2\)).
• Since \(f_1 = f_2\), and \(f = 1/T\), it follows that \(T_1 = T_2\).

Therefore, the time period of the second pendulum must also be 2 seconds. The specific value of the constant phase difference (90°) is not needed to determine the time period, only the fact that it is constant is important.

This demonstrates the fundamental relationship between constant phase difference, frequency, and pendulum time period.

Conclusion

For two pendulums to maintain a constant phase difference during their oscillation, their frequencies must be the same. This directly implies that their time periods must also be the same. Given that one pendulum has a time period of 2 seconds, the other pendulum must also have a time period of 2 seconds.