A and B are two thin rubber bands, each of force constant k. Assuming that they obey Hooke's law, the time period of horizontal SHM of the mass m is given by

2 π \sqrt{\frac{2 m}{k}}

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π \sqrt{\frac{m}{k}}

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2 π \sqrt{\frac{m}{2 k}}

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2 π \sqrt{\frac{m}{k}}

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Solution

The correct option is D $2 π \sqrt{\frac{m}{k}}$
Hooke's law works on the principle of simple harmonic motion.
Similar to SHM, the time period is also same for this rubber band.
Length is replace by mass and the acceleration due to gravity by force constant.
$T = 2 π \sqrt{\frac{m}{k}}$
Option D.

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A and B are two thin rubber bands, each of force constant k. Assuming that they obey Hooke's law, the time period of horizontal SHM of the mass m is given by

The correct option is D 2π√mkHooke's law works on the principle of simple harmonic motion.Similar to SHM, the time period is also same for this rubber band.Length is replace by mass and the acceleration due to gravity by force constant.T=2π√mkOption D.

The correct option is D $2 π \sqrt{\frac{m}{k}}$
Hooke's law works on the principle of simple harmonic motion.
Similar to SHM, the time period is also same for this rubber band.
Length is replace by mass and the acceleration due to gravity by force constant.
$T = 2 π \sqrt{\frac{m}{k}}$
Option D.