1.

A mass of 0.2 kg is attached to the lower end of a massless spring of force constant 200 N m^(-1) , the upper end of which is fixed to a rigid support. Which of the following statement is not true? (Take g=10 m s^(-2))

Answer»

The frequency of oscillation will be nearly 5 Hz.
In EQUILIBRIUM, the spring will be stretched by 2 cm.
If the mass is raised till the spring is in unstretchedstate and then released, it will GO down by 2 cmbefore moving upward.
If the system is taken to a planet, the frequency ofoscillation will be the same as on the earth.

Solution :Frequency of spring , `upsilon=1/(2pi) sqrt(k/m)`
`=1/(2pi) sqrt(200/0.2)`= 5 Hz.
In equilibrium , kx = mg
or `x=(mg)/k=(0.2 xx 10)/200`=0.01 m
When mass is raised till the spring is unstretched, the work
`=1/2kx^2=MGX`
When the mass is released from the unstretched position of spring, then TOTAL work done
mgx.=(mgx)+`1/2kx^2=2mgx`
or x.=2x=2 x 0.1 = 0.02 m
As `upsilon` of spring is INDEPENDENT of g so that the frequency of oscillation will be the same as that on the earth.


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