Question
A 0.2kg. of mass hangs at the end of a spring. When 0.02kg more mass is added to the end of the spring, it stretches 7cm more. If the 0.02kg mass is removed, what will be the period of vibration of the system?
✓
Answer
When 0.02kg is added, there is a stretch of 7cm. Using mg = Kx, we have$\text{K}=\frac{0.02\times10}{7\times10^{-2}}=\frac{20}{7}=2.86\text{N/m}$
Time period $=\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{K}}}$$=2\pi\sqrt{\frac{0.2}{2.86}}=1.66\sec.$
Time period $=\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{K}}}$$=2\pi\sqrt{\frac{0.2}{2.86}}=1.66\sec.$
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