A Spring Stretches By 0.05 M When A Mass Of 0.5 Kg Is Hung From It. A Body Of Mass 1.0 Kg Is Attached To One Of Its Ends, The Other End Being Fixed To The Wall. The Body Is Pulled 0.01 M Along A Horizontal Frictionless Surface And Released. What Is The Total Energy Of The Oscillator. Assume The String To Have Negligible Mass And Take g = 10 Ms–2.  

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Question

A spring stretches by 0.05 m when a mass of 0.5 kg is hung from it. A body of mass 1.0 kg is attached to one of its ends, the other end being fixed to the wall. The body is pulled 0.01 m along a horizontal frictionless surface and released. What is the total energy of the oscillator. Assume the string to have negligible mass and take g = 10 ms–2.  

Solution

Correct option is

0.005 J

 

Force acting on spring = mg = 0.5 × 10 = 5 N. This force extends the spring by 0.05 m. Therefore, the force constant is

                        

The angular frequency of horizontal oscillations is  

                       

This amplitude is A = 0.01 m. Therefore, total energy is

                

SIMILAR QUESTIONS

Q1

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Q2

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Q3

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Q4

 

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Q6

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Q7

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Q8

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Q9

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Q10

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