1.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
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
2. 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
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
3. 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
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.
4. 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
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.
5. 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
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
6. 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
Explanation: V = 2πA/T
V av = 2A/T÷2 = 4A/T
A/T = V/2π
Vav = 2V/π
7. 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
Explanation: V = Aω
<v> = 4A/T
= 2aω/π
= 2V/π
8. 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
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
9. For a body undergoing Simple Harmonic Motion, the acceleration is maximum at the extreme.
a) true
b) false
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
10. 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
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