A particle of mass 10 grams is executing simple harmonic motion with an amplitude of 0-5 m and periodic time of - - 5 seconds- The maximum value of the force acting on the particle is-MP PET 1999- MP PMT 2000-A- 25 NB- 5 NC- 2-5 ND- 0-5 N

The correct option is D 0.5N Maximum force = m( a ω^2 ) = m a ( 4 π^2/T^2) = 0.5 ( 4 π^2/π^2/25) × 0.01 = 0.5 N ...

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Standard XII

Physics

Oscillations

A particle of...

Question

A particle of mass 10 grams is executing simple harmonic motion with an amplitude of 0.5 m and periodic time of (π/5) seconds. The maximum value of the force acting on the particle is

[MP PET 1999; MP PMT 2000]

25N

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2.5N

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0.5N

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Solution

The correct option is D
0.5N

Maximum force = $m (a ω^{2}) = m a (\frac{4 π^{2}}{T^{2}})$

= 0.5 $(\frac{4 π^{2}}{\frac{π^{2}}{25}}) \times 0.01 = 0.5 N$

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A particle of mass 10 grams is executing simple harmonic motion with an amplitude of 0.5 m and periodic time of (π/5) seconds. The maximum value of the force acting on the particle is [MP PET 1999; MP PMT 2000]

The correct option is D 0.5N Maximum force = m(aω2)=ma(4π2T2) = 0.5 (4π2π225)×0.01=0.5N

A particle of mass 10 grams is executing simple harmonic motion with an amplitude of 0.5 m and periodic time of (π/5) seconds. The maximum value of the force acting on the particle is

[MP PET 1999; MP PMT 2000]

The correct option is D
0.5N

Maximum force = $m (a ω^{2}) = m a (\frac{4 π^{2}}{T^{2}})$

= 0.5 $(\frac{4 π^{2}}{\frac{π^{2}}{25}}) \times 0.01 = 0.5 N$