Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?

A

10 m/s^{2}

B

5 m/s^{2}

C

4 m/s^{2}

D

9.8 m/s^{2}

From the potential energy curve,

$${U_{\max }} = {1 \over 2}k{A^2}$$

$$10 = {1 \over 2}k{(2)^2}$$

$$\Rightarrow$$ k = 5 N/m

The length of the simple pendulum, L = 4 m

Time period of spring,

$$T = 2\pi \sqrt {{k \over m}} $$

Time period of simple pendulum,

$$T = 2\pi \sqrt {{l \over g}} $$

The time period of simple pendulum is same as the time period of the spring oscillation.

$$\Rightarrow$$ $$2\pi \sqrt {{l \over g}} $$ = $$2\pi \sqrt {{k \over m}} $$

Substituting the values in the above equations, we get

$$2\pi \sqrt {{4 \over g}} $$ = $$2\pi \sqrt {{5 \over 5}} $$

g = 4m/s^{2}

$$\therefore$$ The acceleration due to the gravity on the planet is 4 m/s^{2}.

$${U_{\max }} = {1 \over 2}k{A^2}$$

$$10 = {1 \over 2}k{(2)^2}$$

$$\Rightarrow$$ k = 5 N/m

The length of the simple pendulum, L = 4 m

Time period of spring,

$$T = 2\pi \sqrt {{k \over m}} $$

Time period of simple pendulum,

$$T = 2\pi \sqrt {{l \over g}} $$

The time period of simple pendulum is same as the time period of the spring oscillation.

$$\Rightarrow$$ $$2\pi \sqrt {{l \over g}} $$ = $$2\pi \sqrt {{k \over m}} $$

Substituting the values in the above equations, we get

$$2\pi \sqrt {{4 \over g}} $$ = $$2\pi \sqrt {{5 \over 5}} $$

g = 4m/s

$$\therefore$$ The acceleration due to the gravity on the planet is 4 m/s

2

MCQ (Single Correct Answer)

For a body executing S.H.M. :

(1) Potential energy is always equal to its K.E.

(2) Average potential and kinetic energy over any given time interval are always equal.

(3) Sum of the kinetic and potential energy at any point of time is constant.

(4) Average K.E. in one time period is equal to average potential energy in one time period.

Choose the most appropriate option from the options given below :

(1) Potential energy is always equal to its K.E.

(2) Average potential and kinetic energy over any given time interval are always equal.

(3) Sum of the kinetic and potential energy at any point of time is constant.

(4) Average K.E. in one time period is equal to average potential energy in one time period.

Choose the most appropriate option from the options given below :

A

(3) and (4)

B

only (3)

C

(2) and (3)

D

only (2)

In S.H.M. total mechanical energy remains constant and also = = $${{1 \over 4}}$$KA^{2} (for 1 time period)

3

MCQ (Single Correct Answer)

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.

The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :

The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :

A

B

C

D

Potential energy is maximum at maximum distance from mean.

4

MCQ (Single Correct Answer)

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $$t = {T \over 4}s$$ starting from mean position. Assume that the initial phase of the oscillation is zero.

A

0.62 J

B

6.2 $$\times$$ 10^{$$-$$3} J

C

1.2 $$\times$$ 10^{3} J

D

6.2 $$\times$$ 10^{3} J

$$T = 2\pi \sqrt {{m \over k}} $$

$$0.2 = 2\pi \sqrt {{{0.5} \over k}} $$

k = 50$$\pi$$^{2}

$$ \approx $$ 500

x = A sin ($$\omega$$t + $$\phi$$)

= 5 cm sin $$\left( {{{\omega T} \over 4} + 0} \right)$$

= 5 cm sin $$\left( {{\pi \over 2}} \right)$$

= 5 cm

$$PE = {1 \over 2}k{x^2}$$

$$ = {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}$$

= 0.6255

$$0.2 = 2\pi \sqrt {{{0.5} \over k}} $$

k = 50$$\pi$$

$$ \approx $$ 500

x = A sin ($$\omega$$t + $$\phi$$)

= 5 cm sin $$\left( {{{\omega T} \over 4} + 0} \right)$$

= 5 cm sin $$\left( {{\pi \over 2}} \right)$$

= 5 cm

$$PE = {1 \over 2}k{x^2}$$

$$ = {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}$$

= 0.6255

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