Question
A particle of mass $0.1kg$ is held between two rigid supports by two springs of force constants $8N/ m$ and $2N/ m$. If the particle is displaced along the direction of the length of the springs, calculate its frequency of vibration.
✓
Answer
The situation is shown in the fig.
When the mass is displaced along the direction of the length of the spring, one spring is compressed while the other is extended but the force due to both the springs is in the same direction. Hence the effective force constant $k = k_1 + k_2 = 8N/ m + 2N/ m = 10N/ m$ The frequency of vibration is given by$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$
$=\frac{1}{2\pi}\sqrt{\frac{10}{0.1}}$
$\text{v}=\frac{10}{2\pi}$
$=\frac{5}{\pi}\text{s}^{-1}$
When the mass is displaced along the direction of the length of the spring, one spring is compressed while the other is extended but the force due to both the springs is in the same direction. Hence the effective force constant $k = k_1 + k_2 = 8N/ m + 2N/ m = 10N/ m$ The frequency of vibration is given by$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$
$=\frac{1}{2\pi}\sqrt{\frac{10}{0.1}}$
$\text{v}=\frac{10}{2\pi}$
$=\frac{5}{\pi}\text{s}^{-1}$
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