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

The correct answer is option (1) 0.00992948m

Step-by-step 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

Answer 2

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

The correct option is A

Step-by-step 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{(m)/(k) }`

Whee m is the particle and k is the spring constant

making
`(m)/(k)` the subject

`(m)/(k) = [(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

`(m)/(k) = (x)/(g)`

Substituting the equation above for
`(m)/(k)`

`[(T)/(2 \pi) ]^2 = (x)/(g)`

making x the subject

`x = g [(T)/(2 \pi) ]`

Substituting the value

`x = 9.8 * [(0.2)/(2 * 3.142) ]^2`

`x = 0.00992948`