Question

A particle hangs from a spring and oscillates with a period of 0.2s.If the mass-spring system remained at rest, by how much would themass stretch it fromits normal equilibrium position? The acceleration of gravity is 9.8m/s2.1. 0.00992948 m2. 0.0397179 m3. 0.019859 m4. 0.0311944 m5. 0.0794358 m6. 0.158872 m

Answer 1

Answer:

The distance the mass would stretch it is    $x = 0.00992948$

The correct option is A

Explanation:

From the question we are told that

           The period of the slit is $T = 0.2s$

           The acceleration due to gravity is $g =9.8 m/s^2$

Generally the period is mathematically represented as

                     $T = 2 \pi \sqrt{\frac{m}{k} }$

          Whee m is the particle and k is the spring constant

        making $\frac{m}{k}$ the subject

                        $\frac{m}{k} = [\frac{T}{2 \pi} ]^2$

The weight on the particle is related to the force stretching the spring by this mathematical relation

               $W = F_s$

              $mg = kx$

where x is the length by which the spring is stretched

          $\frac{m}{k} = \frac{x}{g}$

Substituting the equation above for $\frac{m}{k}$

            $[\frac{T}{2 \pi} ]^2 = \frac{x}{g}$

making x the subject

              $x = g [\frac{T}{2 \pi} ]$

Substituting the value

            $x = 9.8 * [\frac{0.2}{2 * 3.142} ]^2$

            $x = 0.00992948$

Answer 2

Answer:

The correct answer is option (1)  0.00992948m

Explanation:

Given data;

T = 0.2s

g = 9.8mls

The period of oscillation is given by the formula;

T = 2π√m/k

making M/k subject formula, we have

m/k = (T/2π)²

Substituting, we have

m/k = (0.2/2π)²

m/k = 0.0010129

At equilibrium, mg = kx

Making x subject, we have

x = m/k *g

Substituting, we have

x = 0.00101 * 9.8

x   = 0.00992948m