1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :
A
$${A \over 2}$$
B
$${A \over {2\sqrt 2 }}$$
C
$${A \over {\sqrt 2 }}$$
D
A

Explanation

Total energy of particle = $${1 \over 2}k{A^2}$$

Potential energy (v) = $${1 \over 2}$$ kx2

Kinetic energy (K) = $${1 \over 2}$$ kA2 $$-$$ $${1 \over 2}$$kx2

According to the question,

Potential energy = Kinetic energy

$$ \therefore $$  $${1 \over 2}$$kx2 = $${1 \over 2}$$kA2 $$-$$ $${1 \over 2}$$ kx2

$$ \Rightarrow $$  kx2 = $${1 \over 2}$$ kA2

$$ \Rightarrow $$  x = $$ \pm $$ $${A \over {\sqrt 2 }}$$
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz)
A
4
B
7
C
6
D
5

Explanation

For closed organ pipe, resonate frequency is odd multiple of fundamental frequency.

$$ \therefore $$  (2n + 1) f0 $$ \le $$ 20,000

(f0 is fundamental frequency = 1.5 KHz)

$$ \therefore $$  n = 6

$$ \therefore $$  Total number of overtone that can be heared is 7. (0 to 6)
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $$\omega $$. If the radius of the bottle is 2.5 cm then $$\omega $$ is close to – (density of water = 103 kg/m3).
A
2.50 rad s$$-$$1
B
3.75 rad s$$-$$1
C
5.00 rad s$$-$$1
D
7.90 rad s$$-$$1

Explanation

Restoring force due to pressing the bottle with small amount x,

F = $$ - \left( {\rho Ax} \right)g$$

$$ \Rightarrow $$ ma = $$ - \left( {\rho Ax} \right)g$$

$$ \Rightarrow $$ a = $$ - \left( {{{\rho Ag} \over m}} \right)x$$

$$ \therefore $$ $${{\omega ^2} = {{\rho Ag} \over m}}$$ = $${{{\rho \left( {\pi {r^2}} \right)g} \over m}}$$

$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{{{10}^3} \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2} \times 10} \over {310 \times {{10}^{ - 3}}}}} $$ = 7.90 rad/s
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -
A
$${{4\pi } \over 3}$$
B
$${3 \over 8}\pi $$
C
$${7 \over 3}\pi $$
D
$${{8\pi } \over 3}$$

Explanation

$$v = \omega \sqrt {{A^2} - {x^2}} \,\,$$    . . .(1)

$$a = - {\omega ^2}x$$               . . .(2)

$$\left| v \right| = \left| a \right|$$                   . . .(3)

$$\omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x$$

$${A^2} - {x^2} = {\omega ^2}{x^2}$$

$${5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)$$

$$ \Rightarrow \,\,\,3 = \omega \times 4$$

$$T = 2\pi /\omega $$

Questions Asked from Simple Harmonic Motion

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