Search any question & find its solution
Question:
Answered & Verified by Expert
A body of mass $64 \mathrm{~g}$ is made to oscillate turn by turn on two different springs A and B. Spring $\mathrm{A}$ and $\mathrm{B}$ has force constant $4 \frac{\mathrm{N}}{\mathrm{m}}$ and $16 \frac{\mathrm{N}}{\mathrm{m}}$ respectively. If $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ are period of oscillations of springs $\mathrm{A}$ and $\mathrm{B}$ respectively then $\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}$ will be
Options:
Solution:
2936 Upvotes
Verified Answer
The correct answer is:
$3: 1$
(A)
$T=2 \pi \sqrt{\frac{m}{k}} \quad \therefore \frac{T_{1}}{T_{2}}=\sqrt{\frac{k_{2}}{k_{1}}}=\sqrt{\frac{16}{4}}=\frac{2}{1}$
$=\frac{T_{1}+T_{2}}{T_{1}-T_{2}}$
$T=2 \pi \sqrt{\frac{m}{k}} \quad \therefore \frac{T_{1}}{T_{2}}=\sqrt{\frac{k_{2}}{k_{1}}}=\sqrt{\frac{16}{4}}=\frac{2}{1}$
$=\frac{T_{1}+T_{2}}{T_{1}-T_{2}}$
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.