Home » mcq » Civil engineering » Applied mechanics and graphic statics » The piston of a steam engine moves with a simple harmonic motion. The crank rotates 120 r.p.m. and the stroke length is 2 meters. The linear velocity of the piston when it is at a distance of 0.5 metre from the centre, is A. 5.88 m/sec B. 8.88 m/sec C. 10.88 m/sec D. 12.88 m/sec
5.88 m/sec
8.88 m/sec
10.88 m/sec
12.88 m/sec
Answer is Right!
Answer is Wrong!
The correct answer is $\boxed{\text{B) 8.88 m/sec}}$.
The linear velocity of a simple harmonic motion is given
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by the equation $v = \omega \sqrt{A^2 – x^2}$, where $\omega$ is the angular velocity, $A$ is the amplitude, and $x$ is the displacement from the equilibrium position. In this case, we know that $\omega = 2\pi \times \frac{120}{60} = 4\pi$ radians per second, $A = 2$ meters, and $x = 0.5$ meters. Substituting these values into the equation, we get $v = 4\pi \sqrt{(2)^2 – (0.5)^2} = 8.88$ meters per second.
Option A is incorrect because it is the linear velocity of the piston when it is at a distance of 1 meter from the center, which is not the given condition. Option C is incorrect because it is the linear velocity of the piston when it is at a distance of 1.5 meters from the center, which is also not the given condition. Option D is incorrect because it is the linear velocity of the piston when it is at a distance of 2 meters from the center, which is the maximum linear velocity of the piston and is not the given condition.