Answer:
The distance the mass would stretch it is [tex]x = 0.00992948[/tex]
The correct option is A
Explanation:
From the question we are told that
The period of the slit is [tex]T = 0.2s[/tex]
The acceleration due to gravity is [tex]g =9.8 m/s^2[/tex]
Generally the period is mathematically represented as
[tex]T = 2 \pi \sqrt{\frac{m}{k} }[/tex]
Whee m is the particle and k is the spring constant
making [tex]\frac{m}{k}[/tex] the subject
[tex]\frac{m}{k} = [\frac{T}{2 \pi} ]^2[/tex]
The weight on the particle is related to the force stretching the spring by this mathematical relation
[tex]W = F_s[/tex]
[tex]mg = kx[/tex]
where x is the length by which the spring is stretched
[tex]\frac{m}{k} = \frac{x}{g}[/tex]
Substituting the equation above for [tex]\frac{m}{k}[/tex]
[tex][\frac{T}{2 \pi} ]^2 = \frac{x}{g}[/tex]
making x the subject
[tex]x = g [\frac{T}{2 \pi} ][/tex]
Substituting the value
[tex]x = 9.8 * [\frac{0.2}{2 * 3.142} ]^2[/tex]
[tex]x = 0.00992948[/tex]
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
