A mass of 0.2kg is attached to the lower end of a massless spring of force constant 200Nm−1, the upper end of which is fixed to a rigid support. Which of the following statement is true?
A
In equilibrium, the spring will be stretched by 1cm
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B
If the body is raised till the spring attains its natural length and then released, it will go down by 2cm before moving upwards
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C
The frequency of oscillation will be nearly 50Hz
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D
All of the above
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Solution
The correct option is D All of the above Given, Mass of body (m)=0.2kg force constant (K)=200N/m At equilibrium, mg=Kx ⇒x=mgK=6.2×10200=0.01m=1cm Frequency of oscillation (F)=12π√Km =12π√2000.2 =5Hz At equilibrium, the spring is 1cm below the unstretched position. If it is raised to that position and released, let d be displacement of mass before moving up mgd=12Kd2 d=2mgK=2cm spring will go 2cm below the equilibrium position.