Question

# A spring block system is kept on a frictionless surface. Match the respective entries of column (I) with column (II) assuming minimum potential energy for the system as zero. (K = Spring constant, A = Amplitude, m = Mass of block) Column-IColumn-II(A) If mass of the block is doubled(p) Time period increases(keeping K, A unchanged)(B) If the amplitude of oscillation is doubled(q) Time period decreases(keeping K, m unchanged)(C) If force constant is doubled(r) Energy of oscillation increases(keeping m, A unchanged)(D) If another spring of same force constant(s) Energy of oscillation decreasesis attached parallel to the previous one(keeping m, A unchanged)(t) Energy of oscillation remain constant(A)→(r,t); (B)→(r,t); (C)→(p,q); (D)→(q,t)(A)→(s); (B)→(p,q); (C)→(q,r); (D)→(p,t)(A)→(p,t); (B)→(r); (C)→(q,r); (D)→(q,r)(A)→(q,t); (B)→(s); (C)→(p,q); (D)→(q,r)

Solution

## The correct option is C (A)→(p,t); (B)→(r); (C)→(q,r); (D)→(q,r)Since, time period is  T=2π√mK and energy of oscillation, E=12KA2 Options of column: (A)  K and A are constant ⇒E = constant Since, m↑   ∴T↑ ∴(A)→(p,t) (B)  K and m are constant. ∴T = constant Since, A↑   ∴E↑ ∴(B)→(r) (C)  m and A are constant. Since, K↑   ∴T↓ and E↑ ∴(C)→(q,r) (D) Due to attaching another spring K in parallel to previous one. Keq=K+K=2K ⇒K↑ and (m,A) are constant. ∴T↓ and E↑ ∴(D)→(q,r) Thus, option (c) is the correct answer.

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