- Home
- Standard 11
- Physics
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 $g = 10\, ms^{-2}$)
$20$
$10$
$60$
$40$
Solution
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,
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$
$\boldsymbol{\alpha}, \quad \frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$
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
$=\mathrm{Mg}+\mathrm{Kx}+\mathrm{mg}$
$=4 \times 10+625 \times 1.6 \times 10^{-2}+1 \times 10$
$=40+10+10$
$=60 N$
