Question

The scale of a spring balance reading from 0 to 10 kg is 0.25m long. A body suspended from the balance oscillates vertically with a period of \[\dfrac{\pi }{{10}}\] s. The mass suspended is neglected then the mass of the spring is:
(A) 10
(B) 0.98
(C) 5
(D) 20

Answer 1

Hint As the system oscillates with a fixed time period, force of gravity should be equal to elastic force generated by the spring of the spring balance. The spring constant is calculated using \[K = \dfrac{{mg}}{x}\] . Then use the time period formula relating mass, time period and the spring constant. Substitute the values and simplify to get the value of mass.

Complete Step-by-step solution
When the body is suspended from the balance, an elastic force is generated by the spring which is equal to:
F= kx
F= Elastic force
K= spring constant
X= displacement of the spring from mean position
This elastic force will be equal to the force acting on the body due to gravity, which is written as:
F= mg
When the body of mass 10kg is suspended, it stretches the spring by a distance of 0.25m. This will be state of equilibrium and force caused due to gravity and force due to elasticity of spring should be equal
Kx= mg
K= \[\dfrac{{mg}}{x}\]
Where m= mass of the body= 10 kg
 g=acceleration due to gravity = 9.8 \[m{s^2}\]
x- distance moved by the spring = 0.25m
k= \[\dfrac{{10 \times 9.8}}{{0.25}}\]
For an oscillating body, Time period is given by:
T= \[2\pi \sqrt {\dfrac{m}{k}} \]
Where m= mass of oscillating body
k= spring constant
In this question time period is given as \[\dfrac{\pi }{{10}}\]
  \[\therefore \dfrac{\pi }{{10}} = 2\pi \sqrt {\dfrac{m}{k}} \]
 \[\dfrac{1}{{10}} = 2\sqrt {\dfrac{{m \times 0.25}}{{10 \times 9.8}}} \]
Squaring both sides:
\[\dfrac{1}{{{{20}^2}}} = \dfrac{{m \times 0.25}}{{10 \times 9.8}}\]
 \[\dfrac{{10 \times 9.8}}{{{{20}^2} \times 0.25}} = m\]

m= 0.98 kg

Note In this case no damping coefficient or damping force was given, therefore it is advised to ignore it. If in case damping coefficient was given, the equation would have been
 \[2\pi \sqrt {\dfrac{k}{m} - \dfrac{{{b^{^2}}}}{{4{m^2}}}} \]