250+ TOP MCQs on Differential Equation of Simple Harmonic Motion and Answers

Tricky Machine Kinematics Questions and Answers on “Differential Equation of Simple Harmonic Motion”.

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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
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: a
Clarification: 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
Answer: a
Clarification: 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
Answer: a
Clarification: 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
Answer: a
Clarification: 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
Answer: a
Clarification: 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,
T_p = (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
Answer: b
Clarification: 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
A_max = 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
Answer: a
Clarification: 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
Answer: a
Clarification: V = Aω
= 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
Answer: a
Clarification: 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
Answer: a
Clarification: 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
Answer: a
Clarification: 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

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