Question

the pistons in an internal combustion engine undergo a motion that is approximately simple harmonic. if the amplitude of motion is 3.5 cm, and the engine runs at 1700 rev/min, find the maximum acceleration of the pistons.

a. 5.3 km/s²
b. 9.5 km/s²
c. 8.6 km/s²
d. 2.8 km/s²
e. 1.1 km/s²

Explanation:

Step1: Convert revolutions per minute to radians per second

The angular frequency $\omega$ is related to the rotational speed $n$ (in rev/min) by $\omega = 2\pi n/60$. Given $n = 1700$ rev/min, then $\omega=\frac{2\pi\times1700}{60}\text{ rad/s}\approx 178\text{ rad/s}$.

Step2: Recall the formula for maximum acceleration in SHM

The maximum acceleration $a_{max}$ in simple - harmonic motion is given by $a_{max}=\omega^{2}A$, where $A$ is the amplitude. The amplitude $A = 3.5\text{ cm}=0.035\text{ m}$.

Step3: Calculate the maximum acceleration

Substitute $\omega\approx178\text{ rad/s}$ and $A = 0.035\text{ m}$ into the formula $a_{max}=\omega^{2}A$. So $a_{max}=(178)^{2}\times0.035\text{ m/s}^2= 178^{2}\times0.035\text{ m/s}^2=1109.38\text{ m/s}^2\approx1.1\times10^{3}\text{ m/s}^2 = 1.1\text{ km/s}^2$.

Answer:

e. 1.1 km/s²