A mass of 0.2,kg is attached to lower end of a massless spring of force-constant 200, N/m, upper end

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13.Oscillations

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A mass of $0.2\,kg$ is attached to the lower end of a massless spring of force-constant $200\, N/m,$ the upper end of which is fixed to a rigid support. Which of the following statements is/are true ?

A

In equilibrium, the spring will be stretched by $1\,cm.$

B

If the mass is raised till the spring is unstretched state and then released, it will go down by $2\,cm$ before moving upwards.

C

The frequency of oscillation will be nearly $5\, Hz.$

D

all of the above

Solution

The equilibrium stretching is given by equating weight with spring force.

$\therefore 0.2 \times 10=200 \times x$

$x=0.01 m=1 \mathrm{cm}$

When taken back to unstretched state, the sum of spring force and weight together is converted into spring force.

$\therefore 200 \times 0.01+0.2 \times 10=200 \times x$

$x=0.02 m=2 \mathrm{cm}$

The frequency is given by

$f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{200}{0.2}} \approx 5.05 H z$

Standard 11

Physics

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