JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2023
- Medium
- Numerical
In the figure given below. a block of mass \(M =490\,g\) placed on a frictionless table is connected with two springs having same spring constant \(\left( K =2 N m ^{-1}\right)\). If the block is horizontally displaced through ' \(X\) 'm then the number of complete oscillations it will make in \(14 \pi\) seconds will be \(.........\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
\(Keff = K + K\) as both springs are in use in parallel
\(=2\,k\)
\(=2 \times 2=4\,N / m \quad m =490\,gm\)
\(=0.49\,kg\)
\(T =2 \pi \sqrt{\frac{ m }{ Keff }}=2 \pi \sqrt{\frac{0.49\,kg }{4}}\)
\(=2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10}\)
No. of oscillation in the \(14 \pi\) is
\(N =\frac{\text { time }}{ T }=\frac{14 \pi}{7 \pi / 10}=20\)
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