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 \(g = 10\, ms^{-2}\))

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,

\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)

\(\boldsymbol{\alpha}, \quad \frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}\)

or, \(\quad \frac{1}{25}=\sqrt{\frac{1}{\mathrm{K}}}[\because \mathrm{m}=1 \mathrm{kg} ; \omega=25 \mathrm{rad} / \mathrm{s}]\)

or, \(\quad \mathrm{K}=625 \mathrm{Nm}^{-1}\)

The maximum force exerted by the system on the floor

\(=\mathrm{Mg}+\mathrm{Kx}+\mathrm{mg}\)

\(=4 \times 10+625 \times 1.6 \times 10^{-2}+1 \times 10\)

\(=40+10+10\)

\(=60 N\)