Question

Time period of spring-mass, string system
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Answer 1

Given info : A block of mass ‘m’ is attached to spring on one side and elastic string on the other side of constant K.

To find : the time period of spring mass system is...

solution : you see, there are two different things attached to block and block able to move both sides so we have to find time period in both cases individually and then effective time period will be average of them.

case 1 : when block is displaced x distance towards spring.

then, string will elongate so it has tension. and that is Kx. and spring compresses so spring force is Kx. both tension and spring force acts in same direction as you can see in figure.

so, Fnet = -(Kx + Kx)= -2Kx

now, a = -mω²x = -2Kx

⇒ω² = 2K/m

⇒T₁ = 2π√(m/2K) ...(1)

case 2 : when block is displaced x distance towards string.

then, string compresses but compression of string doesn't provide tension on it. so force due to string is zero. but spring elongates so spring force = Kx

so, Fnet = -Kx

so time period in this case, T₂ = 2π√(m/K) = 2√2π√(m/2K)

now, T = (T₁ + T₂)/2

= [2π√(m/2K) + 2√2π√(m/2K) ]/2

= (√2 + 1)π√(m/2K)

Therefore the time period of spring mass string system is (√2 + 1)π√(m/2K)