Another pendulum swings to and fro at a regular rate of
two times per second. Show that its period is 0.5 s.
T=1/f=1/2= 0.5 s
To show that the period of a pendulum that swings at a rate of two times per second is 0.5 seconds, we need to understand the relationship between frequency and period.
The frequency of a pendulum is the number of full oscillations it completes in one second. In this case, the given pendulum swings two times per second, so its frequency is 2 Hz (hertz).
The period of a pendulum is the time it takes to complete one full oscillation. It is calculated by dividing the reciprocal of the frequency. Mathematically, the formula is:
Period (T) = 1 / Frequency (f)
Substituting the given frequency of 2 Hz into the formula:
T = 1 / 2
T = 0.5 s
Thus, we conclude that the period of the pendulum that swings two times per second is 0.5 seconds.
To show that the period of a pendulum that swings at a rate of two times per second is 0.5 seconds, we need to understand the definition of a period and how it relates to the frequency of a pendulum.
The period of a pendulum is the time it takes for one complete cycle, that is, the time it takes for the pendulum to swing from one extreme (e.g., from left to right) and back to the same extreme again. It is usually denoted by the symbol T.
On the other hand, the frequency of a pendulum is the number of complete cycles it completes in one second. It is usually denoted by the symbol f.
The relationship between period and frequency is given by the equation:
frequency (f) = 1 / period (T)
Let's apply this equation to the given scenario:
Given that the pendulum swings at a rate of two times per second, we can conclude that its frequency is 2 Hz (hertz).
Using the equation above, we can solve for the period:
2 Hz = 1 / T
To find the period, we isolate T:
T = 1 / 2 Hz
Simplifying the equation, we have:
T = 0.5 s
Therefore, we have shown that the period of the given pendulum that swings at a rate of two times per second is 0.5 seconds.