A body of mass 100 gram is tied to a spring of spring constant 8 N / m , while the other end of a spring is fixed. If the body moves in a circular path on smooth horizontal surface with constant angul

Question

A body of mass 100 gram is tied to a spring of spring constant 8 N / m , while the other end of a spring is fixed. If the body moves in a circular path on smooth horizontal surface with constant angular speed 8 rad / s then the ratio of extension in the spring to its natured length will be

TestHub · Accepted Answer

Let the natural length of the spring be $L$ and the extension in the spring be $x$ .

The radius of the circular path is $r = L + x$ .

The centripetal force required for circular motion is provided by the restoring force of the spring.

$m ω^{2} r = k x$

$m ω^{2} (L + x) = k x$

Given $m = 100 g = 0.1 kg$ , $k = 8 N/m$ , and $ω = 8 rad/s$ . Substituting these values:

$0.1 \times (8)^{2} \times (L + x) = 8 x$

$0.1 \times 64 \times (L + x) = 8 x$

$6.4 (L + x) = 8 x$

$6.4 L + 6.4 x = 8 x$

$8 x - 6.4 x = 6.4 L$

$1.6 x = 6.4 L$

$\frac{x}{L} = \frac{6.4}{1.6} = 4$

The ratio of extension in the spring to its natural length is $4 : 1$ .

Answer: $4 : 1$

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