Engineering Questions with Answers - Multiple Choice Questions

Home » MCQs » Aeronautical Engineering » Machine Kinematics – Differential Equation of Simple Harmonic Motion

# Machine Kinematics – Differential Equation of Simple Harmonic Motion

1 - Question

By how much angle in degrees does the velocity leads the displacement in a body undergoing SIMPLE HARMONIC MOTION?

a) 90

b) 45

c) 180

d) 0

**
View Answer**

Answer: aExplanation: For a body undergoing Simple Harmonic Motion, the velocity leads the displacement by an angle of 90 degrees as shown by the differential equation of the motion.

2 - Question

For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the direction of the displacement.

a) True

b) False

**
View Answer**

Answer: bExplanation: For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the opposite direction of the displacement. This is indicated by a negative sign in the equation.

3 - Question

The maximum displacement of the body under Simple Harmonic Motion from its mean position is known as_______

a) Amplitude

b) Frequency

c) Time period

d) Range

**
View Answer**

Answer: aExplanation: The maximum displacement of a body undergoing Simple Harmonic Motion is known as Amplitude, it is generally denoted by ‘A’.

4 - Question

The piston of an engine moves with SIMPLE HARMONIC MOTION. The crank rotates at a speed of 120 r.p.m. with a stroke of 2 metres. Find the velocity of the piston in m/s, when it is at a distance of 0.75 metre from the centre.

a) 8.31

b) 7.33

c) 8.41

d) 9.02

**
View Answer**

Answer: aExplanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation v=ω(A2−x2)−−−−−−−−√ substituting the values we get v = 8.31 m/s.

5 - Question

The piston of an engine moves with SIMPLE HARMONIC MOTION. The crank rotates at a speed of 120 r.p.m. with a stroke of 2 metres. Find the velocity of the piston in m/s2, when it is at a distance of 0.75 metre from the centre.

a) 118.46

b) 117.33

c) 128.41

d) 119.02

**
View Answer**

Answer: aExplanation: When a body is undergoing SIMPLE HARMONIC MOTION, its acceleration is given by the equation a=ω2x substituting the values we get a = 118.46 m/s2

6 - Question

A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its angular velocity in rad/s.

a) 5.7

b) 7.5

c) 6.7

d) 7.6

**
View Answer**

Answer: aExplanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation v = ωr2–x2−−−−−√ Given values: when point (x) = 0.75 m, then its velocity (v) = 11 m/s when point (x) = 2 m, then its velocity (v) = 3 m/s When point is 0.75 m away from the mid path (v) is, v = ωr2–x2−−−−−√ 11 = ωr2–(0.75)2−−−−−−−−√ —>Eq(1) Similarly, When point is 2 m away from the centre (v) is, v = ωr2–x2−−−−−√ 3 = ωr2–22−−−−−√ —>Eq(2) Solving Eq(1) & Eq(2) we get, 113ωr2–(0.75)2√ωr2–22√=r2–(0.75)2√r2–22√ Squaring on both sides, we get 1219=r2–0.5625r2–4 121r2 – 484 = 9r2 – 5.0625 121r2 – 9r2 = 484 – 5.0625 112r2 = 478.9375 r2 = 478.9375112 r2 = 4.27622 r = 2.07m Substituting the value of r in Eq(1) we get, 11 = ω(2.07)2–(0.75)2−−−−−−−−−−−−√ 11 = ω4.2849–0.5625−−−−−−−−−−−−√ 11 = ω3.7224−−−−−√ 11 = 1.9293 ω ω = 111.9293 = 5.7 rad/s.

7 - Question

A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its time period in s.

a) 1.1

b) 1.2

c) 1.3

d) 1.4

**
View Answer**

Answer: aExplanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation v = ωr2–x2−−−−−√ Given values: when point (x) = 0.75 m, then its velocity (v) = 11 m/s when point (x) = 2 m, then its velocity (v) = 3 m/s When point is 0.75 m away from the mid path (v) is, v = ωr2–x2−−−−−√ 11 = ωr2–(0.75)2−−−−−−−−√ —>Eq(1) Similarly, When point is 2 m away from the centre (v) is, v = ωr2–x2−−−−−√ 3 = ωr2–22−−−−−√ —>Eq(2) Solving Eq(1) & Eq(2) we get, 113ωr2–(0.75)2√ωr2–22√=r2–(0.75)2√r2–22√ Squaring on both sides, we get 1219=r2–0.5625r2–4 121r2 – 484 = 9r2 – 5.0625 121r2 – 9r2 = 484 – 5.0625 112r2 = 478.9375 r2 = 478.9375112 r2 = 4.27622 r = 2.07m Substituting the value of r in Eq(1) we get, 11 = ω(2.07)2–(0.75)2−−−−−−−−−−−−√ 11 = ω4.2849–0.5625−−−−−−−−−−−−√ 11 = ω3.7224−−−−−√ 11 = 1.9293 ω ω = 111.9293 = 5.7 rad/s. We know periodic time is, Tp = 2πω=2π5.7=2×3.145.7=6.285.7 = 1.1 s.

8 - Question

A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its maximum acceleration in m/s2.

a) 61.1

b) 67.2

c) 51.3

d) 41.4

**
View Answer**

Answer: bExplanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation v = ωr2–x2−−−−−√ Given values: when point (x) = 0.75 m, then its velocity (v) = 11 m/s when point (x) = 2 m, then its velocity (v) = 3 m/s When point is 0.75 m away from the mid path (v) is, v = ωr2–x2−−−−−√ 11 = ωr2–(0.75)2−−−−−−−−√ —>Eq(1) Similarly, When point is 2 m away from the centre (v) is, v = ωr2–x2−−−−−√ 3 = ωr2–22−−−−−√ —>Eq(2) Solving Eq(1) & Eq(2) we get, 113ωr2–(0.75)2√ωr2–22√=r2–(0.75)2√r2–22√ Squaring on both sides, we get 1219=r2–0.5625r2–4 121r2 – 484 = 9r2 – 5.0625 121r2 – 9r2 = 484 – 5.0625 112r2 = 478.9375 r2 = 478.9375112 r2 = 4.27622 r = 2.07m Substituting the value of r in Eq(1) we get, 11 = ω(2.07)2–(0.75)2−−−−−−−−−−−−√ 11 = ω4.2849–0.5625−−−−−−−−−−−−√ 11 = ω3.7224−−−−−√ 11 = 1.9293 ω ω = 111.9293 = 5.7 rad/s. Maximum acceleration is Amax = ω2r = (5.7)2 × 2.07 = 32.49 × 2.07 = 67.25 m/s2.

9 - Question

If V is the maximum velocity of a body undergoing SIMPLE HARMONIC MOTION, then what is the average velocity of motion from one extreme to other extreme is?

a) 2V/π

b) 4V/π

c) V/2π

d) 2V/3π

**
View Answer**

Answer: aExplanation: V = 2πA/T V av = 2A/T÷2 = 4A/T A/T = V/2π Vav = 2V/π

10 - Question

If V is the maximum velocity of a body undergoing SIMPLE HARMONIC MOTION, then what is the average velocity of motion?

a) 2V/π

b) 4V/π

c) V/2π

d) 2V/3π

**
View Answer**

Answer: aExplanation: V = Aω = 4A/T = 2aω/π = 2V/π

11 - Question

Which of the following is the correct differential equation of the SIMPLE HARMONIC MOTION?

a) d2x/dt2 + ω2x = 0

b) d2x/dt2 – ω2x = 0

c) d2x/dt + ω2x = 0

d) d2x/dt – ω2x = 0

**
View Answer**

Answer: aExplanation: For body undergoing Simple Harmonic Motion, it’s motion can be represented as projected uniform circular motion with radius equal to the amplitude of motion. Therefore x =Acosωt dx/dt = -Aωsinωt d2x/dt2 = -aω2cosωt therefore d2x/dt2 + ω2x = 0

12 - Question

For a body undergoing Simple Harmonic Motion, the acceleration is maximum at the extreme.

a) True

b) False

**
View Answer**

Answer: aExplanation: For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the opposite direction of the displacement. This is indicated by a negative sign in the equation and at an extreme position, the acceleration attains a maximum value.

13 - Question

Which of the following is the solution of the differential equation of the SIMPLE HARMONIC MOTION?

a) x = Acosωt + B sinωt

b) x = (A+B)cosωt

c) x = (A+B)sinωt

d) x = Atanωt + B sinωt

**
View Answer**

Answer: aExplanation: We know that dx/dt = -Aωsinωt d2x/dt2 = -aω2cosωt therefore d2x/dt2 + ω2x = 0 is the standard differential equation of Simple Harmonic Motion It’s solution is/are: x = Acosωt + B sinωt