Question

When a load of 0.5 kg is attached to a spring, the extension is 0.2m. If a load of 0.25 kg is attached and released, the time period of oscillation in sec is (Assume g = 10 m/s²)

• A. π/5 sec.
• B. 2π/5 sec
• C. 2π sec.
• D. 20 π

Answer 1

Answer: (A)

π/5 sec.

Answer 2

0.5g = kx

or, k = $\frac { 0.5 \times 10 } { 0.2 }$ = 25 N/m

T = 2π $\sqrt { \frac { 0.25 } { 25 } }$ = $\frac { \pi } { 5 }$sec.