Two bodies of masses 1kg and 4kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25rad/s, and amplitude 1.6cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is $({take g= 10ms}^{-2})$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

20N

60N

10N

40N

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Mass of bigger body M=4kg
Mass of smaller body m=1kg
Smaller mass (m=1kg) executes S.H.M of angular frequency $ω = 25 r a d / s$
Amplitude x=1.6cm= $1.6 \times 10^{- 2}$
As we know,
T= $2 π \sqrt{\frac{m}{K}}$
Or, $\frac{2 π}{ω}$ = $2 π \sqrt{\frac{m}{K}}$
Or, $\frac{1}{25} = \sqrt{\frac{1}{K}}$ $[∵ m = 1 k g; ω = 25 r a d / s]$
Or, K=625 $N m^{- 1}$
The maximum force exerted by the system on the floor
=Mg+Kx+mg
= $4 \times 10 + 625 \times 1.6 \times 10^{- 2} + 1 \times 10$
=60N