Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

Structural Analysis

Construction Material and Management

Reinforced Cement Concrete

Steel Structures

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

Hydrology

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

Two particles are performing simple harmonic motion in a straight line about
the same equilibrium point. The amplitude and time period for both particles are same and equal to A and *I*, respectively. At time t = 0 one particle has
displacement A while the other one has displacement $${{ - A} \over 2}$$ and they are moving towards each other. If they cross each other at time t, then t is :

A

$${T \over 6}$$

B

$${5T \over 6}$$

C

$${T \over 3}$$

D

$${T \over 4}$$

Angular displacement ($$\theta $$

$${y_1}$$ = A sin$$\theta$$

$$ \Rightarrow $$ A = Asin$$\theta $$

$$ \Rightarrow $$ sin$$\theta $$

$$ \therefore $$ $$\theta $$

Similarly for particle 2 angular displacement $$\theta $$

y

$$ \Rightarrow $$ $$-$$ $${A \over 2}$$ = Asin$$\theta $$

$$ \Rightarrow $$ sin$$\theta $$

$$ \Rightarrow $$ $$\theta $$

Relative angular displacement of the two particle,

$$\theta $$ = $$\theta $$

= $${{\pi \over 2}}$$ $$-$$ $$\left( { - {\pi \over 6}} \right)$$

= $${{{2\pi } \over 3}}$$

Relative angular velocity $$=$$ $$\omega - \left( { - \omega } \right)$$ = $$2\omega $$

If they cross each other at time t

then, t = $${\theta \over {2\omega }}$$ = $${{2\pi } \over {3 \times 2\omega }}$$ = $${\pi \over {3 \times {{2\pi } \over T}}}$$ = $${T \over 6}$$

2

In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :

A

0.1 Hz

B

1.2 Hz

C

0.7 Hz

D

1.9 Hz

Here,

Amplitude, A = 7 cm = 0.07 m

When washer is no longer stays in contact with the piston, then the normal force on the washer is = 0

$$ \therefore $$ Maximum acceleration of the washer,

a_{max} = $$\omega $$^{2}A = g

$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{g \over A}} $$ = $$\sqrt {{{10} \over {0.07}}} $$ = $$\sqrt {{{1000} \over 7}} $$

$$ \therefore $$ Frequency of the piston,

f = $${\omega \over {2\pi }}$$ = $${1 \over {2\pi }}\sqrt {{{1000} \over 7}} $$ = 1.9 Hz

Amplitude, A = 7 cm = 0.07 m

When washer is no longer stays in contact with the piston, then the normal force on the washer is = 0

$$ \therefore $$ Maximum acceleration of the washer,

a

$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{g \over A}} $$ = $$\sqrt {{{10} \over {0.07}}} $$ = $$\sqrt {{{1000} \over 7}} $$

$$ \therefore $$ Frequency of the piston,

f = $${\omega \over {2\pi }}$$ = $${1 \over {2\pi }}\sqrt {{{1000} \over 7}} $$ = 1.9 Hz

3

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :

A

$${1 \over 4}Hz$$

B

$${1 \over {2\sqrt 2 }}Hz$$

C

$${1 \over 2}Hz$$

D

$$2$$ $$Hz$$

Here frequency of spring (f) = $${1 \over {2\pi }}\sqrt {{k \over m}} $$

Given that, F = 1 Hz

$$\therefore\,\,\,$$ $${1 \over {2\pi }}\sqrt {{k \over 1}} $$ = 1

$$ \Rightarrow $$$$\,\,\,$$ k = 4$$\pi $$

Here two identical springs are attached in parallel. So,

K

$$\therefore\,\,\,$$ Frequency of 8 kg block,

F' = $${1 \over {2\pi }}\sqrt {{{{k_{eq}}} \over {m'}}} $$

= $${1 \over {2\pi }}\sqrt {{{2k} \over 8}} $$

= $${1 \over {2\pi }}\sqrt {{{2 \times 4{\pi ^2}} \over 8}} $$

= $${1 \over {2\pi }} \times \pi $$

= $${1 \over 2}$$ Hz

4

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s^{−1} . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is $${\pi \over 4}$$.

A

500 m/s^{2}

B

500 $$\sqrt 2 m/$$ s^{2}

C

750 m/s^{2}

D

750 $$\sqrt 2 $$m / s^{2}

Mximum velocity, V_{max} = a$$\omega $$

Maximum acceleration, A_{max} = a$$\omega $$^{2}

Given that,

$${{a{\omega ^2}} \over {a\omega }}$$ = 10

$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = 10 s^{$$-$$1}

Displacement, x = a sin ($$\omega $$t + $${\pi \over 4}$$)

at t = 0, displacement x = 5

$$\therefore\,\,\,$$ 5 = a sin $$\left( {{\pi \over 4}} \right)$$

$$ \Rightarrow $$$$\,\,\,$$ 5 = a $$ \times $$ $${1 \over {\sqrt 2 }}$$

$$ \Rightarrow $$$$\,\,\,$$ a = 5$${\sqrt 2 }$$

$$\therefore\,\,\,$$ Maximum acceleration,

A_{max} = a$$\omega $$^{2} = 5 $${\sqrt 2 }$$ $$ \times $$ (10)^{2}

= 500 $${\sqrt 2 }$$ m/s^{2}

Maximum acceleration, A

Given that,

$${{a{\omega ^2}} \over {a\omega }}$$ = 10

$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = 10 s

Displacement, x = a sin ($$\omega $$t + $${\pi \over 4}$$)

at t = 0, displacement x = 5

$$\therefore\,\,\,$$ 5 = a sin $$\left( {{\pi \over 4}} \right)$$

$$ \Rightarrow $$$$\,\,\,$$ 5 = a $$ \times $$ $${1 \over {\sqrt 2 }}$$

$$ \Rightarrow $$$$\,\,\,$$ a = 5$${\sqrt 2 }$$

$$\therefore\,\,\,$$ Maximum acceleration,

A

= 500 $${\sqrt 2 }$$ m/s

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