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A spring balance has a scale which range...

A spring balance has a scale which ranges from 0 to 25 kg and the length of the scale is 0.25 m. It is taken to an unknown planet X where the acceleration due to gravity is `11.5ms^(-1)`. Suppose a body a mass M kg is suspended in this spring and made to oscillate with a period of 0.50 s. Compute the gravitational force acting on the body.

Text Solution

Verified by Experts

Let us first calculate the stiffness constant of the spring balance by using equation `(m)/(k)=(l)/(g)`
`k=(mg)/(l)=(25xx11.5)/(0.25)=1150Nm^(-1)`
The time period of oscillations is given by `T=2pisqrt((M)/(k))`, where M is the mass of the body.
Since, M is unknown, rearranging, we get
`M=(kT^(2))/(4pi^(2))=((1150)(0.5)^(2))/(4pi^(2))=7.3kg` `M=(kT^(2))/(4pi^(2))=((1150)(0.5)^(2))/(4pi^(2))=7.3kg`
The gravitational force acting on the body is
`W=Mg=7.3xx11.5=83.95N~~84N`
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