Answer to Question #274949 in Physics for Dhea

Question

A particle describes simple harmonic motion with amplitude of 5cm and period of 0.2 sec. Calculate the velocity and acceleration of the particle when the displacement is 5 cm. Calculate the velocity and acceleration of the particle when the displacement is 3 cm.

Answer 1

a)

$$a=-(10\pi)^2(0.05)=-49.3\frac{m}{s^2}\\V=0\frac{m}{s}$$

b)

$$a=-(10\pi)^2(0.03)=-29.6\frac{m}{s^2}\\V=10\pi\sqrt{0.05^2-0.03^2}=1.26\frac{m}{s}$$