One end of a spring is fixed to the ceiling and other end is attached to a block. The block is released when spring is relaxed. The product of time period and amplitude is 8 S.I. units. If spring is cut in two equal parts and the two springs are attached to the block as shown in figure. The block is released when both springs are relaxed. Now find the product of time period and amplitude in S.I. units.

Question

One end of a spring is fixed to the ceiling and other end is attached to a block. The block is released when spring is relaxed. The product of time period and amplitude is 8 S.I. units. If spring is cut in two equal parts and the two springs are attached to the block as shown in figure. The block is released when both springs are relaxed. Now find the product of time period and amplitude in S.I. units. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Now K of spring system changes and become 4 K. Before: K A=m g ⇒ A=(m g/K) ; T=2 π √(m/K) ∵ T A=8 After: (m g/4 K)=(A/4) ⇒ T'=2 π √(m/4 K)=(T/2) ⇒ T' A'=(T A/8)=(8/8)=1

Now K of spring system changes and become 4K. Before: KA=mg ⇒A=Kmg​; T=2πKm​​ ∵TA=8 After: 4Kmg​=4A​ ⇒T′=2π4Km​​=2T​ ⇒T′A′=8TA​=88​=1

Now $K$ of spring system changes and become $4 K$ .
Before: $K A = m g$
$\Rightarrow A = \frac{m g}{K};$
$T = 2 π \frac{m}{K}$
$∵ T A = 8$
After: $\frac{m g}{4 K} = \frac{A}{4}$
$\Rightarrow T^{'} = 2 π \frac{m}{4 K} = \frac{T}{2}$
$\Rightarrow T^{'} A^{'} = \frac{T A}{8} = \frac{8}{8} = 1$