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

Answer:

The the magnitude of the maximum velocity and acceleration are 16.93 m/s and $5.213\times10^{3}\ m/s^2$.

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

$v=A\times\omega$

Where, A = amplitude

Put the value into the formula

$v=5.50\times10^{-2}\times2940\times\dfrac{2\pi}{60}$

$v=16.93\ m/s$

(b). We need to calculate the maximum acceleration

Using formula of maximum acceleration

$a=A\times\omega^2$

$a=5.50\times10^{-2}\times(2940\times\dfrac{2\pi}{60})^2$

$a=5.213\times10^{3}\ m/s^2$

Hence, The the magnitude of the maximum velocity and acceleration are 16.93 m/s and $5.213\times10^{3}\ m/s^2$.