This set of Tricky Machine Kinematics Questions and Answers focuses on “Differential Equation of Simple Harmonic Motion”.
1. 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
Explanation: 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. For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the direction of the displacement.
a) True
b) False
View Answer
Explanation: 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. 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
Explanation: The maximum displacement of a body undergoing Simple Harmonic Motion is known as Amplitude, it is generally denoted by ‘A’.
4. 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
Explanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
\(v = \omega \sqrt{(A^2-x^2)}\)
substituting the values we get
v = 8.31 m/s.
5. 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
Explanation: 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. 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
Explanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = \(\omega \sqrt{r^2 – x^2}\)
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 = \(\omega \sqrt{r^2 – x^2}\)
11 = \(\omega \sqrt{r^2 – (0.75)^2}\) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = \(\omega \sqrt{r^2 – x^2}\)
3 = \(\omega \sqrt{r^2 – 2^2}\) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
\(\frac{11}{3} \frac{\omega \sqrt{r^2 – (0.75)^2}}{\omega \sqrt{r^2 – 2^2}} = \frac{\sqrt{r^2 – (0.75)^2}}{\sqrt{r^2 – 2^2}} \)
Squaring on both sides, we get
\(\frac{121}{9} = \frac{r^2 – 0.5625}{r^2 – 4}\)
121r2 – 484 = 9r2 – 5.0625
121r2 – 9r2 = 484 – 5.0625
112r2 = 478.9375
r2 = \(\frac{478.9375}{112}\)
r2 = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = \( \omega \sqrt{(2.07)^2 – (0.75)^2}\)
11 = \( \omega \sqrt{4.2849 – 0.5625}\)
11 = \( \omega \sqrt{3.7224}\)
11 = 1.9293 ω
ω = \(\frac{11}{1.9293}\) = 5.7 rad/s.
7. 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
Explanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = \(\omega \sqrt{r^2 – x^2}\)
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 = \(\omega \sqrt{r^2 – x^2}\)
11 = \(\omega \sqrt{r^2 – (0.75)^2}\) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = \(\omega \sqrt{r^2 – x^2}\)
3 = \(\omega \sqrt{r^2 – 2^2}\) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
\(\frac{11}{3} \frac{\omega \sqrt{r^2 – (0.75)^2}}{\omega \sqrt{r^2 – 2^2}} = \frac{\sqrt{r^2 – (0.75)^2}}{\sqrt{r^2 – 2^2}} \)
Squaring on both sides, we get
\(\frac{121}{9} = \frac{r^2 – 0.5625}{r^2 – 4}\)
121r2 – 484 = 9r2 – 5.0625
121r2 – 9r2 = 484 – 5.0625
112r2 = 478.9375
r2 = \(\frac{478.9375}{112}\)
r2 = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = \( \omega \sqrt{(2.07)^2 – (0.75)^2}\)
11 = \( \omega \sqrt{4.2849 – 0.5625}\)
11 = \( \omega \sqrt{3.7224}\)
11 = 1.9293 ω
ω = \(\frac{11}{1.9293}\) = 5.7 rad/s.
We know periodic time is,
Tp = \(\frac{2 \pi}{\omega} = \frac{2 \pi}{5.7} = \frac{2 \times 3.14}{5.7} = \frac{6.28}{5.7}\) = 1.1 s.
8. 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
Explanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = \(\omega \sqrt{r^2 – x^2}\)
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 = \(\omega \sqrt{r^2 – x^2}\)
11 = \(\omega \sqrt{r^2 – (0.75)^2}\) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = \(\omega \sqrt{r^2 – x^2}\)
3 = \(\omega \sqrt{r^2 – 2^2}\) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
\(\frac{11}{3} \frac{\omega \sqrt{r^2 – (0.75)^2}}{\omega \sqrt{r^2 – 2^2}} = \frac{\sqrt{r^2 – (0.75)^2}}{\sqrt{r^2 – 2^2}} \)
Squaring on both sides, we get
\(\frac{121}{9} = \frac{r^2 – 0.5625}{r^2 – 4}\)
121r2 – 484 = 9r2 – 5.0625
121r2 – 9r2 = 484 – 5.0625
112r2 = 478.9375
r2 = \(\frac{478.9375}{112}\)
r2 = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = \( \omega \sqrt{(2.07)^2 – (0.75)^2}\)
11 = \( \omega \sqrt{4.2849 – 0.5625}\)
11 = \( \omega \sqrt{3.7224}\)
11 = 1.9293 ω
ω = \(\frac{11}{1.9293}\) = 5.7 rad/s.
Maximum acceleration is
Amax = ω2r = (5.7)2 × 2.07 = 32.49 × 2.07
= 67.25 m/s2.
9. 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
Explanation: V = 2πA/T
V av = 2A/T÷2 = 4A/T
A/T = V/2π
Vav = 2V/π
10. 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
Explanation: V = Aω
<v> = 4A/T
= 2aω/π
= 2V/π
11. 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
Explanation: 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. For a body undergoing Simple Harmonic Motion, the acceleration is maximum at the extreme.
a) True
b) False
View Answer
Explanation: 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. 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
Explanation: 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
Sanfoundry Global Education & Learning Series – Machine Kinematics.
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