The minimum time taken by the spring block system having time period T to travel distance equal to amplitude of motion is:
$\begin{align}
& a)\dfrac{T}{4} \\
& b)\dfrac{T}{6} \\
& c)\dfrac{T}{8} \\
& d)\dfrac{T}{10} \\
\end{align}$
Answer
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Hint: The amplitude of a spring block system is simply the maximum displacement of the object attached to spring from the equilibrium position. Find the distance travelled by the spring for amplitude of the motion. Calculate the time period for the time taken for two extreme positions. The difference between them is the time minimum.
Formula used:
$\begin{align}
& y=A\sin \omega t \\
& \omega =\dfrac{2\pi }{T} \\
\end{align}$
Complete answer:
Let us say x and y are the distance travelled by the spring before coming to rest.
$\begin{align}
& x=\dfrac{r}{2} \\
& y=r \\
\end{align}$
Now,
$\begin{align}
& x=A\sin \omega {{t}_{1}} \\
& \Rightarrow \dfrac{r}{2}=r\sin \omega {{t}_{1}} \\
& \Rightarrow \sin \omega {{t}_{1}}=\dfrac{1}{2} \\
& \Rightarrow \omega {{t}_{1}}=\dfrac{\pi }{6} \\
& \Rightarrow {{t}_{1}}=\dfrac{T}{12} \\
\end{align}$
Similarly, for the second distance travelled,
The time taken will be,
$\begin{align}
& r=r\sin \omega {{t}_{2}} \\
& \Rightarrow \sin \omega {{t}_{2}}=1 \\
& \Rightarrow \omega {{t}_{2}}=\dfrac{\pi }{2} \\
& \Rightarrow {{t}_{2}}=\dfrac{\pi }{2}\times \dfrac{T}{2\pi } \\
& \Rightarrow {{t}_{2}}=\dfrac{T}{4} \\
\end{align}$
As the time taken is the difference between the two equations,
$\begin{align}
& {{t}_{2}}-{{t}_{1}}=\dfrac{T}{4}-\dfrac{T}{12} \\
& \Rightarrow {{t}_{2}}-{{t}_{1}}=\dfrac{T}{6} \\
\end{align}$
So, the correct answer is “Option B”.
Additional Information:
The amplitude is the maximum distance travelled in a single period of time by the spring. Time period is the total time taken by the spring for one complete oscillation. Unit of time is seconds. Amplitude can also be stated as the distance between the resting point and the maximum displacement of the block. Frequency is the number of times the object passes a certain point in a given time. Time period is the time taken for one complete oscillation by the spring block system.
Note:
The period does not depend on the Amplitude. The more amplitude the more distance to cover but the faster it will cover the distance. The distance and speed will cancel each other out, so the period will remain the same.
Formula used:
$\begin{align}
& y=A\sin \omega t \\
& \omega =\dfrac{2\pi }{T} \\
\end{align}$
Complete answer:
Let us say x and y are the distance travelled by the spring before coming to rest.
$\begin{align}
& x=\dfrac{r}{2} \\
& y=r \\
\end{align}$
Now,
$\begin{align}
& x=A\sin \omega {{t}_{1}} \\
& \Rightarrow \dfrac{r}{2}=r\sin \omega {{t}_{1}} \\
& \Rightarrow \sin \omega {{t}_{1}}=\dfrac{1}{2} \\
& \Rightarrow \omega {{t}_{1}}=\dfrac{\pi }{6} \\
& \Rightarrow {{t}_{1}}=\dfrac{T}{12} \\
\end{align}$
Similarly, for the second distance travelled,
The time taken will be,
$\begin{align}
& r=r\sin \omega {{t}_{2}} \\
& \Rightarrow \sin \omega {{t}_{2}}=1 \\
& \Rightarrow \omega {{t}_{2}}=\dfrac{\pi }{2} \\
& \Rightarrow {{t}_{2}}=\dfrac{\pi }{2}\times \dfrac{T}{2\pi } \\
& \Rightarrow {{t}_{2}}=\dfrac{T}{4} \\
\end{align}$
As the time taken is the difference between the two equations,
$\begin{align}
& {{t}_{2}}-{{t}_{1}}=\dfrac{T}{4}-\dfrac{T}{12} \\
& \Rightarrow {{t}_{2}}-{{t}_{1}}=\dfrac{T}{6} \\
\end{align}$
So, the correct answer is “Option B”.
Additional Information:
The amplitude is the maximum distance travelled in a single period of time by the spring. Time period is the total time taken by the spring for one complete oscillation. Unit of time is seconds. Amplitude can also be stated as the distance between the resting point and the maximum displacement of the block. Frequency is the number of times the object passes a certain point in a given time. Time period is the time taken for one complete oscillation by the spring block system.
Note:
The period does not depend on the Amplitude. The more amplitude the more distance to cover but the faster it will cover the distance. The distance and speed will cancel each other out, so the period will remain the same.
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