Answer:
Original length = 2.97 m
Explanation:
Let the original length of the pendulum be 'L' m
Given:
Acceleration due to gravity (g) = 9.8 m/s²
Original time period of the pendulum (T) = 3.45 s
Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.
New length of the pendulum is, [tex]L_1=L-1[/tex]
New time period of the pendulum is, [tex]T_1=2.81\ s[/tex]
We know that, the time period of a simple pendulum of length 'L' is given as:
[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]-------------- (1)
So, for the new length, the time period is given as:
[tex]T_1=2\pi\sqrt{\frac{L_1}{g}}[/tex]------------ (2)
Squaring both the equations and then dividing them, we get:
[tex]\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1[/tex]
Now, plug in the given values and calculate 'L'. This gives,
[tex]L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m[/tex]
Therefore, the original length of the simple pendulum is 2.97 m
