Chapter 3: Problem 45

We know \(\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \Rightarrow \mathrm{T} \Rightarrow \sqrt{\ell}\) \(\mathrm{T}^{2}=4 \pi^{2} \frac{\ell}{\mathrm{g}} \Rightarrow \mathrm{T}^{2} \alpha \ell\) From above relationships, it is clear that as the length of simple pendulum increases, time period also increases and vice versa.

Short Answer

Expert verified

Answer: As the length (l) of the simple pendulum increases, the time period (T) also increases.

Step by step solution

The formula T = 2π√(l/g) is used to find the time period (T) of a simple pendulum, where 'l' is the length of the pendulum and 'g' is the acceleration due to gravity. This formula demonstrates that T is directly proportional to the square root of 'l'. #2. Analyze the alternative formula for T^2 [T^2 = 4π^2(l/g)]#

The alternative formula T^2 = 4π^2(l/g) emphasizes the proportionality between T^2 and 'l'. It's important to note that as 'l' increases, so does T^2. #3. Show that an increase in length 'l' corresponds to an increase in time period 'T'#

Since T^2 is directly proportional to 'l', if we increase the length (l) of the pendulum, the value of T^2 will also increase. As T^2 increases, T will also increase, as T = √(T^2). This demonstrates that as the length of the simple pendulum increases, the time period also increases. #4. Conclusion#

From the given formulas T = 2π√(l/g) and T^2 = 4π^2(l/g), it is clear that as the length 'l' of the simple pendulum increases, so does the time period 'T'. This is due to the direct proportionality between T^2 and 'l', and the relationship that T = √(T^2).