Question

The vertical extension in a light spring by a weight of 1 kg suspended from the wire is 9.8 cm. The period of oscillation

• A. $20 \pi \mathrm { sec }$
• B. $2 \pi \mathrm { sec }$
• C.
• D. $200 \pi \mathrm { sec }$

Answer 1

Answer: (C)

Answer 2

$\bullet \bullet$ mg = kx ⇒ $\frac { m } { k } = \frac { x } { g }$⇒ $T = 2 \pi \sqrt { \frac { m } { k } } = 2 \pi \sqrt { \frac { x } { g } }$

$= 2 \pi \sqrt { \frac { 9.8 \times 10 ^ { - 2 } } { 9.8 } } = \frac { 2 \pi } { 10 }$sec