Question

piston in a gasoline engine is in simple harmonic motion. The engine is running at the rate of 2 940 rev/min. Taking the extremes of its position relative to its center point as ±5.50 cm.(a) Find the magnitude of the maximum velocity of the piston. 16.93 Correct: Your answer is correct. m/s(b) Find the magnitude of the maximum acceleration of the piston

Answer 1

The the magnitude of the maximum velocity and acceleration are 16.93 m/s and
`5.213*10^(3)\ m/s^2`.

Step-by-step explanation:

Given that,

Rate = 2940 rev/min

Amplitude = ±5.50 cm

(a). We need to calculate the magnitude of the maximum velocity

Using formula of maximum velocity

`v=A*\omega`

Where, A = amplitude

Put the value into the formula

`v=5.50*10^(-2)*2940*(2\pi)/(60)`

`v=16.93\ m/s`

(b). We need to calculate the maximum acceleration

Using formula of maximum acceleration

`a=A*\omega^2`

`a=5.50*10^(-2)*(2940*(2\pi)/(60))^2`

`a=5.213*10^(3)\ m/s^2`

Hence, The the magnitude of the maximum velocity and acceleration are 16.93 m/s and
`5.213*10^(3)\ m/s^2`.