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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 KA unchanged)(B) If the amplitude of oscillation is doubled(q) Time period decreases(keeping Km unchanged)(C) If force constant is doubled(r) Energy of oscillation increases(keeping mA unchanged)(D) If another spring of same force constant(s) Energy of oscillation decreasesis attached parallel to the previous one(keeping mA unchanged)(t) Energy of oscillation remain constant
  1. (A)(r,t); (B)(r,t); (C)(p,q); (D)(q,t)
  2. (A)(s); (B)(p,q); (C)(q,r); (D)(p,t)
  3. (A)(p,t); (B)(r); (C)(q,r); (D)(q,r)
  4. (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=2πmK
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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