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Sagot :
The period T of a pendulum with length L is given by the formula:
[tex]T=2\pi\cdot\sqrt[]{\frac{L}{g}}[/tex]Where g is the acceleration of gravity:
[tex]g=9.81\frac{m}{s^2}[/tex]If the length of the pendulum is doubled, the period will be:
[tex]\begin{gathered} T^{\prime}=2\pi\cdot\sqrt[]{\frac{2L}{g}} \\ =\sqrt[]{2}\times2\pi\cdot\sqrt[]{\frac{L}{g}} \\ =\sqrt[]{2}\times T \end{gathered}[/tex]On the other hand, the frequency is the reciprocal of the period. Then:
[tex]\begin{gathered} f=\frac{1}{T} \\ f^{\prime}=\frac{1}{T^{\prime}}=\frac{1}{\sqrt[]{2}\times T}=\frac{1}{\sqrt[]{2}}\times\frac{1}{T}=\frac{1}{\sqrt[]{2}}\times f \end{gathered}[/tex]Then, if the length of the pendulum is doubled, the frequency is cut by a factor of 1/√2:
[tex]f^{\prime}=\frac{1}{\sqrt[]{2}}\times f[/tex]This is not the same as if the frequency was cut by a factor of 1/2.
Therefore, the statement is:
[tex]\text{False}[/tex]
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