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

Question

A particle of mass 10 g is executing simple harmonic motion with an amplitude of 0.5 m and periodic time of ( π /5 ) second. The maximum value of the force acting on the particle is (A) 25 N (B) 5N (C) 2.5 N (D) 0.5 N

Tardigrade Admin · Accepted Answer

Maximum force F=mω 2a =m( (2π /T) )2a=(4π 2/T2)ma =(4π 2/( (π )5 )2)× (10/1000)× 0.5 =0.5 text N

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

25 N

5N

2.5 N

0.5 N

Maximum force $F = m ω^{2} a$ $= m (\frac{2 π}{T})^{2} a = \frac{4 π ^{2}}{T ^{2}} ma$ $= \frac{4 π ^{2}}{( \frac{π}{5} ) ^{2}} \times \frac{10}{1000} \times 0.5$ $= 0.5 N$

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

25 N

5N

2.5 N

0.5 N

Maximum force F=mω2a =m(T2π​)2a=T24π2​ma =(5π​)24π2​×100010​×0.5 =0.5N

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

Maximum force $F = m ω^{2} a$ $= m (\frac{2 π}{T})^{2} a = \frac{4 π ^{2}}{T ^{2}} ma$ $= \frac{4 π ^{2}}{( \frac{π}{5} ) ^{2}} \times \frac{10}{1000} \times 0.5$ $= 0.5 N$