Question

A block of mass m rigidly attached with a spring of stiffness k is compressed through a distance A. If the block is released, the period of oscillation of the block for a complete cycle is equal to

• A.
• B. $\frac { \pi } { \sqrt { 2 } } \sqrt { \frac { \mathrm { m } } { \mathrm { k } } }$
• C. $\frac { 2 \pi } { 3 } \sqrt { \frac { \mathrm { m } } { \mathrm { k } } }$
• D. None of these.

Answer 1

Answer: (A)

Answer 2

The period of motion from A to is equal to quarter of the time period T of oscillation of m-k system.

$\Rightarrow$ $\mathrm { t } _ { \mathrm { AO } } = \frac { \mathrm { T } } { 4 } = \frac { 1 } { 4 } \left[ 2 \pi \sqrt { \frac { \mathrm { m } } { \mathrm { k } } } \right] = \frac { \pi } { 2 } \sqrt { \frac { \mathrm { m } } { \mathrm { k } } }$

Since the motion is simple harmonic

OB = OA sin where is the time of motion from O to B.

$\Rightarrow$ $\mathrm { t } _ { \mathrm { OB } } = \frac { \mathrm { T } } { 2 \pi } \sin ^ { - 1 } \frac { \mathrm {~A} / 2 } { \mathrm {~A} } = \frac { \mathrm { T } } { 2 \pi } \left( \frac { \pi } { 6 } \right) = \frac { \mathrm { T } } { 12 } = \frac { 2 \pi } { 12 } \sqrt { \frac { \mathrm {~m} } { \mathrm { k } } } = \frac { \pi } { 6 } \sqrt { \frac { \mathrm {~m} } { \mathrm { k } } }$

$\therefore$ The total time of motion for a complete cycle = t = 2 $\left( \mathrm { t } _ { \mathrm { AO } } + \mathrm { t } _ { \mathrm { OB } } \right)$

$\Rightarrow$ $\mathrm { t } = 2 \left[ \frac { \pi } { 2 } \sqrt { \frac { \mathrm { m } } { \mathrm { k } } } + \frac { \pi } { 6 } \sqrt { \frac { \mathrm { m } } { \mathrm { k } } } \right] = \frac { 4 \pi } { 3 } \sqrt { \frac { \mathrm { m } } { \mathrm { k } } }$

Hence correct choice is : (1)