Search any question & find its solution
Question:
Answered & Verified by Expert
A particle of mass $m$ is attached to three identical massless springs of spring constant ' $k$ ' as shown in the figure. The time period of vertical oscillation of the particle is

Options:

Solution:
1858 Upvotes
Verified Answer
The correct answer is:
$2 \pi \sqrt{\frac{m}{2 k}}$
$$
\begin{aligned}
& \text { Hints : } \mathrm{T}=2 \pi \sqrt{\frac{m}{\mathrm{~K}_{\mathrm{eq}}}} \\
& \mathrm{F}=\mathrm{K} x+2 \mathrm{~K} x \cos ^2 45 \\
& \mathrm{~K}_{\mathrm{eq}} x=\mathrm{K} x+\mathrm{K} x \\
& \mathrm{~K}_{\mathrm{eq}}=2 \mathrm{~K}
\end{aligned}
$$
\begin{aligned}
& \text { Hints : } \mathrm{T}=2 \pi \sqrt{\frac{m}{\mathrm{~K}_{\mathrm{eq}}}} \\
& \mathrm{F}=\mathrm{K} x+2 \mathrm{~K} x \cos ^2 45 \\
& \mathrm{~K}_{\mathrm{eq}} x=\mathrm{K} x+\mathrm{K} x \\
& \mathrm{~K}_{\mathrm{eq}}=2 \mathrm{~K}
\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.