JEE Mains · Physics · STD 11 - 13. oscillations
Two bodies of masses \(1\, kg\) and \(4\, kg\) are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency \(25\, rad/s\), and amplitude \(1.6\, cm\) while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is ..... \(N\) ( take \(g = 10\, ms^{-2}\))
- A \(20\)
- B \(10\)
- C \(60\)
- D \(40\)
Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
Mass of bigger body \(M=4 \mathrm{kg}\) Mass of smaller body \(\mathrm{m}=1 \mathrm{kg}\) Smaller mass \((\mathrm{m}=1 \mathrm{kg})\) \(executes \,S.H.M \,of\) angular frequency \(\omega=25\) rad \(/ \mathrm{s}\) Amplitude \(x=1.6 \mathrm{cm}=1.6 \times 10^{-2}\) As we know,…
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