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”.

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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