To solve this problem we will proceed to define the Period of a stick, then we will define the frequency, which is the inverse of the period. We will compare the change suffered by the new length and replace that value. The Time period of meter stick is
[tex]T = 2\pi\sqrt{ \frac{L}{g}}[/tex]
Here,
L = Length
g = Gravity
At the same time the frequency is
[tex]f = \frac{1}{T}[/tex]
Therefore the frequency in Terms of the Period is
[tex]f_0 = \frac{1}{2\pi}\sqrt{\frac{g}{L}}[/tex]
If bottom third were cut off then the new length is
[tex]L' = \frac{2}{3} L[/tex]
Replacing this value at the new frequency we have that,
[tex]f ' = \frac{1}{2\pi} \sqrt{\frac{3g}{2L}}[/tex]
[tex]f' = \sqrt{\frac{3}{2}}(\frac{1}{2\pi}\sqrt{\frac{g}{L}})[/tex]
Finally,
[tex]\therefore f' = \sqrt{\frac{3}{2}}f_0[/tex]
Answer: f = √2(f0)
Explanation:
For a swing, the frequency function can be written as
f0 = 1/2π √(mgl/i)
Where f0 = frequency, m = mass, g = acceleration due to gravity,
Therefore for the length
f0 = 1/2π√(g/L)
Where L = length
When the length is cut by half, L = L/2
Substituting L with L/2 we have;
f = 1/2π√(g/(L/2))
f = 1/2π√(2g/L)
f = √2/2π√(2g/L)
f = √2×f0
f = √2(f0)
