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) 5 N (C) 2.5 N (D) 0.5 N

Tardigrade Admin · Accepted Answer

Force in a simple harmonic motion F=m f=m(-ω2 x)=-m ω2 A sin ω t For maximum value of F, sin ω t=1 ∴ F max =m ω2 A =m((2 π/T))2 A =(m 4 π2 A/T2) =(10 × 10-3 × 4 π2 × 0.5/π2 / 25)=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

5 N

2.5 N

0.5 N

Force in a simple harmonic motion
$F = m f = m (- ω^{2} x) = - m ω^{2} A sin ω t$
For maximum value of $F, sin ω t = 1$
$∴ F_{m a x} = m ω^{2} A$
$= m (\frac{2 π}{T})^{2} A = \frac{m 4 π ^{2} A}{T ^{2}}$
$= \frac{10 \times 1 0 ^{- 3} \times 4 π ^{2} \times 0.5}{π ^{2} /25} = 0.5 N$

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

25 N

5 N

2.5 N

0.5 N

Force in a simple harmonic motion F=mf=m(−ω2x)=−mω2Asinωt For maximum value of F,sinωt=1 ∴Fmax​=mω2A =m(T2π​)2A=T2m4π2A​ =π2/2510×10−3×4π2×0.5​=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

Force in a simple harmonic motion
$F = m f = m (- ω^{2} x) = - m ω^{2} A sin ω t$
For maximum value of $F, sin ω t = 1$
$∴ F_{m a x} = m ω^{2} A$
$= m (\frac{2 π}{T})^{2} A = \frac{m 4 π ^{2} A}{T ^{2}}$
$= \frac{10 \times 1 0 ^{- 3} \times 4 π ^{2} \times 0.5}{π ^{2} /25} = 0.5 N$