
When the length of a simple pendulum is decreased by 20cm, the period changes by \[10\%\]. Find the original length of the pendulum.
Answer
567.6k+ views
Hint: We are given the length of the pendulum at an instant, which was changed from the original value. The rate of change in period as observed is also given. We can easily substitute this in the formula for time period and get the original length of the pendulum.
Complete answer:
We know that the length of a pendulum is directly related to its time taken for an oscillation.
We know that we get the relation for the time period of oscillation of a pendulum and the length of the pendulum by solving the equation of simple harmonic motion. We know that the time period is directly proportional to the square root of the length of pendulum as –
\[T\propto \sqrt{\dfrac{l}{g}}\]
Where, T is the time period of oscillation,
l is the length of pendulum
g is the acceleration due to gravity.
We get the equation for time period as –
\[T=2\pi \sqrt{\dfrac{l}{g}}\]
Now, let us substitute the given data in this equation. Let us assume the original length of the pendulum as ‘l’ and the new length as ‘l’’, which is given as –
\[l'=l-20cm\]
And let the initial time period be ‘T’ and the new time period be ‘T’’ given as –
\[T'=T-\dfrac{10}{100}T=0.9T\]
Now, let us find the time period for the second condition as –
\[\begin{align}
& T'=2\pi \sqrt{\dfrac{l'}{g}} \\
& \Rightarrow \text{ }T'=2\pi \sqrt{\dfrac{l-20cm}{g}} \\
\end{align}\]
But the initial time period T and T’ are related as –
\[T'=0.9T\]
From this we can derive the length as –
\[\begin{align}
& T'=2\pi \sqrt{\dfrac{l'}{g}} \\
& \Rightarrow \text{ (0}\text{.9)}2\pi \sqrt{\dfrac{l}{g}}=2\pi \sqrt{\dfrac{l-20cm}{g}} \\
& \Rightarrow \text{ }0.9\sqrt{l}=\sqrt{l-0.2} \\
& \Rightarrow \text{ 0}\text{.81}l=l-0.2 \\
& \Rightarrow \text{ 0}\text{.19}l=0.2 \\
& \therefore \text{ }l=1.053m \\
\end{align}\]
The actual length of the pendulum was 1.053m.
Note:
The units need to be treated carefully while calculating. Also, a percentage change indicates the change from its initial condition and therefore, should be subtracted from the initial value to get the intended value of the quantity under consideration.
Complete answer:
We know that the length of a pendulum is directly related to its time taken for an oscillation.
We know that we get the relation for the time period of oscillation of a pendulum and the length of the pendulum by solving the equation of simple harmonic motion. We know that the time period is directly proportional to the square root of the length of pendulum as –
\[T\propto \sqrt{\dfrac{l}{g}}\]
Where, T is the time period of oscillation,
l is the length of pendulum
g is the acceleration due to gravity.
We get the equation for time period as –
\[T=2\pi \sqrt{\dfrac{l}{g}}\]
Now, let us substitute the given data in this equation. Let us assume the original length of the pendulum as ‘l’ and the new length as ‘l’’, which is given as –
\[l'=l-20cm\]
And let the initial time period be ‘T’ and the new time period be ‘T’’ given as –
\[T'=T-\dfrac{10}{100}T=0.9T\]
Now, let us find the time period for the second condition as –
\[\begin{align}
& T'=2\pi \sqrt{\dfrac{l'}{g}} \\
& \Rightarrow \text{ }T'=2\pi \sqrt{\dfrac{l-20cm}{g}} \\
\end{align}\]
But the initial time period T and T’ are related as –
\[T'=0.9T\]
From this we can derive the length as –
\[\begin{align}
& T'=2\pi \sqrt{\dfrac{l'}{g}} \\
& \Rightarrow \text{ (0}\text{.9)}2\pi \sqrt{\dfrac{l}{g}}=2\pi \sqrt{\dfrac{l-20cm}{g}} \\
& \Rightarrow \text{ }0.9\sqrt{l}=\sqrt{l-0.2} \\
& \Rightarrow \text{ 0}\text{.81}l=l-0.2 \\
& \Rightarrow \text{ 0}\text{.19}l=0.2 \\
& \therefore \text{ }l=1.053m \\
\end{align}\]
The actual length of the pendulum was 1.053m.
Note:
The units need to be treated carefully while calculating. Also, a percentage change indicates the change from its initial condition and therefore, should be subtracted from the initial value to get the intended value of the quantity under consideration.
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