Question

21. Two springs of force-constants 100 N mand 400 Nm are connected to a mass 5 kg, as shown What is the time period of horizontal oscillations? Derive the expression used,​

Answer 1

The equivalent spring constant of two springs in series is given as,

K=K1+K2K1K2=100×300/400=75 N/m

So the time period for SHM of the given spring-mass system is,

T=2πm/k=2π3/75=2π/5=(2π×20)/(5×20)=40π/100

I hope so this answer