14.24 A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.

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Vishal Baghel | Contributor-Level 10

5 months ago

Amplitude = 5 cm = 0.05 m

Time period, T = 0.2 s

(a) For displacement x = 5 cm = 0.05m

Acceleration is given by a = ${- ω}^{2} x$ = $- ({\frac{2 π}{T})}^{2} x$ = $- ({\frac{2 π}{0.2})}^{2} \times 0.05$ = $-$ 5 $π^{2}$ m/s

Velocity is given by V = $ω \sqrt{A^{2} - x^{2}}$ = $\frac{2 π}{T} \sqrt{{0.05}^{2} - {0.05}^{2}}$ = 0

(b) For displacement x = 3 cm = 0.03m

Acceleration is given by a = ${- ω}^{2} x$ = $- ({\frac{2 π}{T})}^{2} x$ = $- ({\frac{2 π}{0.2})}^{2} \times 0.03$ = $-$ 3 $π^{2}$ m/s

Velocity is given by V = $ω \sqrt{A^{2} - x^{2}}$ = $\frac{2 π}{T} \sqrt{{0.05}^{2} - {0.03}^{2}}$ = 0.4 $π m / s$

(c) For displacement x = 0 cm = 0 m

Acceleration is given by a = ${- ω}^{2} x$ = $- ({\frac{2 π}{T})}^{2} x$ = $- ({\frac{2 π}{0.2})}^{2} \times 0$ = $0$ m/s

Velocity is given by V = $ω \sqrt{A^{2} - x^{2}}$ = $\frac{2 π}{T} \sqrt{{0.05}^{2} 0}$ = 0.5 $π m / s$

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