Answer:
The maximum speed can the child walk is v/sqrt 2
Explanation:
Let the length of parent’s leg be L
Then the length of the child’s leg [tex]=\frac{L}{2}[/tex]
Maximum Speed at which the parent walks=V
To Find :
The maximum speed at which the child walks=?
Solution:
The Frequency of the simple pendulum
[tex]f=\frac{1}{2 \pi}(\sqrt{\frac{g}{l}})[/tex]
[tex]\text {Speed}=\frac{\text {distance}}{\text {time}}[/tex]
Speed of the parent
[tex]V=L f_{p}[/tex]
[tex]V=L\left(\frac{1}{2 \pi}(\sqrt{\frac{g}{L}})\right)[/tex]
Speed of the child
[tex]v=\frac{L}{2}\left(\frac{1}{2 \pi}(\sqrt{\frac{g}{L / 2}})\right)[/tex]
Now,
[tex]\frac{v}{V}=\frac{\frac{L}{2}\left(\frac{1}{2 \pi}(\sqrt{\frac{g}{L / 2}})\right)}{L\left(\frac{1}{2 \pi}(\sqrt{\frac{g}{L}})\right)}[/tex]
[tex]\frac{v}{V}=\frac{\sqrt{2}}{2}[/tex]
[tex]\frac{v}{V}=\frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}[/tex]
[tex]\frac{v}{V}=\frac{1}{\sqrt{2}}[/tex]
[tex]v=\frac{V}{\sqrt{2}}[/tex]
Result:
The maximum Speed at which the child can walk is[tex]\frac{V}{\sqrt{2}}[/tex]
