By suspending a mass of 0.50 kg, a spring is stretched by 8.20 m. If a mass of 0.25 kg is suspended, then its period of oscillation will be :
(Take g = 10 $m s^{- 2}$ )

0.137 s

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0.328 s

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0.628 s

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1.000 s

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Solution

The correct option is C 0.628 s
The force constant of the spring,
$k = \frac{F}{x} = \frac{0.5 \times 10}{0.2} = 25 N / m$
Now, if the mass of 0.25 kg is suspended by the spring, then the period of oscillation,
$T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{0.25}{25}} = 0.628 s$

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By suspending a mass of 0.50 kg, a spring is stretched by 8.20 m. If a mass of 0.25 kg is suspended, then its period of oscillation will be :(Take g = 10 ms−2)

The correct option is C 0.628 sThe force constant of the spring,k=Fx=0.5×100.2=25N/mNow, if the mass of 0.25 kg is suspended by the spring, then the period of oscillation,T=2π√mk=2π√0.2525=0.628s

By suspending a mass of 0.50 kg, a spring is stretched by 8.20 m. If a mass of 0.25 kg is suspended, then its period of oscillation will be :
(Take g = 10 $m s^{- 2}$ )

The correct option is C 0.628 s
The force constant of the spring,
$k = \frac{F}{x} = \frac{0.5 \times 10}{0.2} = 25 N / m$
Now, if the mass of 0.25 kg is suspended by the spring, then the period of oscillation,
$T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{0.25}{25}} = 0.628 s$