Search any question & find its solution
Question:
Answered & Verified by Expert
A spring has a natural length $l$ with one end fixed to the ceiling. The other end is fitted with a smooth ring which can slide on a horizontal rod fixed at distance $l$ below the ceiling.
Initially, the spring makes an angle of $60^{\circ}$ with the vertical, when system is released from rest. Find the angle of the spring with the vertical, when the velocity of the ring reaches half of the maximum velocity, which the ring can attain during the motion.
Options:
Initially, the spring makes an angle of $60^{\circ}$ with the vertical, when system is released from rest. Find the angle of the spring with the vertical, when the velocity of the ring reaches half of the maximum velocity, which the ring can attain during the motion.
Solution:
1289 Upvotes
Verified Answer
The correct answer is:
None of the above
$$
\text { From diagram, }
$$
$$
\cos 60^{\circ}=\frac{l}{h} \Rightarrow h=\frac{l}{\cos 60^{\circ}} \Rightarrow h=2 l
$$
Extension in spring $=h-l=2 l-l=l$
At mean position, if velocity is $v$ and there is no friction, then
$$
\frac{1}{2} m v^2=\frac{1}{2} k x^2 \Rightarrow v^2=\frac{k}{m} l^2
$$
If angle with vertical is $\theta$ when velocity is $v^{\prime}=\frac{v}{2}$
Then, $\quad \frac{1}{2} m v^{\prime 2}=\frac{1}{2} k x^{\prime 2} \quad \frac{1}{2} m\left(\frac{v}{2}\right)^2=\frac{1}{2} k\left(\frac{l}{\cos \theta^{\prime}}-l\right)^2$
As, $\quad \cos \theta^{\prime}=\frac{l}{h^{\prime}} \quad \Rightarrow \quad h^{\prime}=\frac{l}{\cos \theta^{\prime}}$ and extension, $X^{\prime}=h^{\prime}-l=\frac{l}{\cos \theta^{\prime}}-l$
Substituting value of $v^2$, we have
$$
\begin{aligned}
& \frac{1}{2} m\left(\frac{h l^2}{4 m}\right)=\frac{1}{2} k\left(\frac{l}{\cos \theta^{\prime}}-l\right)^2 \\
& \Rightarrow \frac{l}{2}=\frac{l}{\cos \theta^{\prime}}-l \Rightarrow \frac{3 l}{2}=\frac{l}{\cos \theta^{\prime}} \\
& \cos \theta^{\prime}=\frac{2}{3} \quad \theta^{\prime}=\cos ^{-1} \frac{2}{3}
\end{aligned}
$$
(No option is matching).
\text { From diagram, }
$$
$$
\cos 60^{\circ}=\frac{l}{h} \Rightarrow h=\frac{l}{\cos 60^{\circ}} \Rightarrow h=2 l
$$
Extension in spring $=h-l=2 l-l=l$
At mean position, if velocity is $v$ and there is no friction, then
$$
\frac{1}{2} m v^2=\frac{1}{2} k x^2 \Rightarrow v^2=\frac{k}{m} l^2
$$
If angle with vertical is $\theta$ when velocity is $v^{\prime}=\frac{v}{2}$
Then, $\quad \frac{1}{2} m v^{\prime 2}=\frac{1}{2} k x^{\prime 2} \quad \frac{1}{2} m\left(\frac{v}{2}\right)^2=\frac{1}{2} k\left(\frac{l}{\cos \theta^{\prime}}-l\right)^2$
As, $\quad \cos \theta^{\prime}=\frac{l}{h^{\prime}} \quad \Rightarrow \quad h^{\prime}=\frac{l}{\cos \theta^{\prime}}$ and extension, $X^{\prime}=h^{\prime}-l=\frac{l}{\cos \theta^{\prime}}-l$
Substituting value of $v^2$, we have
$$
\begin{aligned}
& \frac{1}{2} m\left(\frac{h l^2}{4 m}\right)=\frac{1}{2} k\left(\frac{l}{\cos \theta^{\prime}}-l\right)^2 \\
& \Rightarrow \frac{l}{2}=\frac{l}{\cos \theta^{\prime}}-l \Rightarrow \frac{3 l}{2}=\frac{l}{\cos \theta^{\prime}} \\
& \cos \theta^{\prime}=\frac{2}{3} \quad \theta^{\prime}=\cos ^{-1} \frac{2}{3}
\end{aligned}
$$
(No option is matching).
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.