Search any question & find its solution
Question:
Answered & Verified by Expert
A particle is suspended from a vertical spring which is executing S.H.M. of frequency $5 \mathrm{~Hz}$. The spring is unstretched at the highest point of oscillation. Maximum speed of the particle is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$
Options:
Solution:
1787 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{\pi} \mathrm{m} / \mathrm{s}$
Amplitude of SHM, $\mathrm{A}=\frac{\mathrm{mg}}{\mathrm{k}}=\frac{\mathrm{g}}{\omega^2}$
$$
\begin{aligned}
& \omega=2 \pi \mathrm{f}=2 \pi \times 5=10 \pi \\
& \therefore \mathrm{A}=\frac{10}{100 \pi^2}=\frac{1}{10 \pi^2} \\
& \mathrm{~V}_{\max }=\mathrm{A} \omega=\frac{1}{10 \pi^2} \times 10 \pi=\frac{1}{\pi} \mathrm{m} / \mathrm{s}
\end{aligned}
$$
$$
\begin{aligned}
& \omega=2 \pi \mathrm{f}=2 \pi \times 5=10 \pi \\
& \therefore \mathrm{A}=\frac{10}{100 \pi^2}=\frac{1}{10 \pi^2} \\
& \mathrm{~V}_{\max }=\mathrm{A} \omega=\frac{1}{10 \pi^2} \times 10 \pi=\frac{1}{\pi} \mathrm{m} / \mathrm{s}
\end{aligned}
$$
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.