JEE Mains · Physics · STD 11 - 13. oscillations
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 \(.........\)
- A \(20\)
- B \(21\)
- C \(19\)
- D \(26\)
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…
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