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### 1.2.9 Example

A body undergoes a simple harmonic motion with amplitude 3.0 m and period 4.00 s. Find the

a) angular frequency,

b) maximum speed,

c) maximum acceleration,

d) acceleration of the body when it is 1.0 m from the equilibrium point, and

e) speed of the body when it is 1.0 m from the equilibrium point.

[ 1.57 rad/s, 4.7 m/s , 7.4 m/s^2 , -2.5 m/s^2, 4.4 m/s ]

## Hint:

### 1.2.9.1 Solution:

a) angular frequency , ω = $\genfrac{}{}{0.1ex}{}{2\pi}{T}=\genfrac{}{}{0.1ex}{}{2\pi}{4}=1.57\genfrac{}{}{0.1ex}{}{rad}{s}$

b) maximum speed, v_{0}= x

_{0}ω = (3.00)(1.57) = 0.047 m/s

c) maximum acceleration, |a

_{0}| = x

_{0}ω

^{2}= (3.00)(1.57)

^{2}= 7.4 m/s

^{2}

d) when x = 1.00, acceleration, a = -ω

^{2}x = (1.57)

^{2}(1.00) = -2.5 m/s

^{2}

e) when x = 1.00, speed, v = x

_{0}ω = (3.00)(1.57) = 4.4 m/s

### 1.2.9.2 Model:

### For Teachers

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