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.

Asked by Abhisek | 1 year ago |  148

1 Answer

Solution :-

Amplitude = 5 cm = 0.05 m

Time period = 0.2 s

When the displacement is y, then

acceleration, A = -ω2y

Velocity, v =$$\omega \sqrt{r^{2}-y^{2}}$$

ω = $$\dfrac{2π}{T}$$

$$\dfrac{2π}{0.2}$$= 10π rad/s

(a) When the displacement y = 5 cm = 0.05 m

Acceleration, A = – (10π)2(0.05)

= 5π2 m/s2

Velocity, V = $$10\pi \sqrt{(0.05)^{2}-(0.05)^{2}}=0$$

(b) When the displacement y = 3 cm = 0.03 m

Acceleration, A = – (10π)2(0.03)

= 3π2 m/s2

Velocity, V = $$10\pi \sqrt{(0.05)^{2}-(0.03)^{2}}$$

$$=10\pi \times 0.04 = 0.4\pi m/s$$

(c) When the displacement y = 0

Acceleration, A =  – (10π)2(0) = 0

Velocity, V = $$10\pi \sqrt{(0.05)^{2}-(0)^{2}}$$

$$=10\pi \times 0.05 = 0.5\pi m/s$$

Answered by Abhisek | 1 year ago

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