Question

Match the column I with column II

	Column I		Column II
(A)		(p)	$\mathrm { T } = 2 \pi \sqrt { \frac { \mathrm {~m} \left( \mathrm { k } _ { 1 } + \mathrm { k } _ { 2 } \right) } { \mathrm { k } _ { 1 } \mathrm { k } _ { 2 } } }$
(B)		(q)	$\mathrm { T } = 2 \pi \sqrt { \frac { 2 \mathrm {~m} } { \mathrm { k } } }$
(C)		(r)	$\mathrm { T } = 2 \pi \sqrt { \frac { \mathrm {~m} } { 2 \mathrm { k } } }$
(D)		(s)	$\mathrm { T } = 2 \pi \sqrt { \frac { \mathrm {~m} } { \mathrm { k } _ { 1 } + \mathrm { k } _ { 2 } } }$

• A. A-p, B-q, C-s, D-r
• B. A-s, B-r, C-p, D-q
• C. A-r, B-p, C-s, D-q
• D. A-p, B-r, C-q, D-s

Answer 1

Answer: (B)

A-s, B-r, C-p, D-q

Answer 2

In figure A, two springs are connected in parallel the effective spring constant is

$\therefore \mathrm { T } = 2 \pi \sqrt { \frac { \mathrm {~m} } { \mathrm { k } _ { 1 } + \mathrm { k } _ { 2 } } }$

A – s

In figure B two identical springs are connected in parallel. the effective spring constant is

B –r

In figure C, two springs are connected in series the effective spring’s constant is

Or

$\therefore \mathrm { T } = 2 \pi \sqrt { \frac { \mathrm {~m} \left( \mathrm { k } _ { 1 } + \mathrm { k } _ { 2 } \right) } { \mathrm { k } _ { 1 } \mathrm { k } _ { 2 } } }$

C – p

In figure D, two identical sprigs are connected in series the effective spring constant is

$\mathrm { k } _ { \text {eff } } = \frac { ( \mathrm { k } ) ( \mathrm { k } ) } { \mathrm { k } + \mathrm { k } } = \frac { \mathrm { k } } { 2 }$

$\therefore \mathrm { T } = 2 \pi \sqrt { \frac { 2 \mathrm {~m} } { \mathrm { k } } }$

D – q