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1

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)
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
2

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)
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

JEE Main 2019 (Online) 9th January Evening Slot

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

एक कण $$x$$-अक्ष की दिशा में, $$x=0$$ के सापेक्ष आयाम $$\mathrm{A}$$ से सरल आवर्त गति कर रहा है। जब इस कण की स्थितिज ऊर्जा तथा गतिज ऊर्जा के मान बराबर हैं, तो कण की स्थिति होगी :

A
$$\frac{\mathrm{A}}{2}$$
B
$$\frac{\mathrm{A}}{2 \sqrt{2}}$$
C
$$\frac{\mathrm{A}}{\sqrt{2}}$$
D
$$\mathrm{A}$$
4

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)
English
Hindi
A particle executes simple harmonic motion and is located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The freqquency of the oscillation is :
A
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)$$
B
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)$$
C
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)$$
D
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)$$

Explanation

In general equation of simple harmonic motion, y = A sin $$\omega $$t

$$\therefore\,\,\,$$ a = A sin $$\omega $$t0

$$\,\,\,\,\,\,$$ b = A sin 2$$\omega $$t0

$$\,\,\,\,\,\,\,$$c = A sin 3$$\omega $$t0

a + c = A[sin $$\omega $$t0 + sin 3$$\omega $$t0]

= 2A sin 2$$\omega $$t0 cos$$\omega $$t0

$$ \Rightarrow $$$$\,\,\,$$a + c = 2 b cos$$\omega $$t0

$$ \Rightarrow $$$$\,\,\,$$ $${{a + c} \over b}$$ = 2 cos$$\omega $$t0

$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = $${1 \over {{t_0}}}$$ cos$$-$$1 $$\left( {{{a + c} \over {2b}}} \right)$$

$$\therefore\,\,\,$$ f = $${\omega \over {2\pi }}$$

= $${1 \over {2\pi {t_0}}}$$ cos$$-$$1 $$\left( {{{a + c} \over {2b}}} \right)$$

एक कण सरल आवर्त गति करता है और समय $$\mathrm{t}_{0}$$, $$2 \mathrm{t}_{\mathrm{0}}$$ तथा $$3 \mathrm{t}_{\mathrm{0}}$$ पर उसकी स्थिति क्रमश: $$x=\mathrm{a}, \mathrm{b}$$ तथा $$\mathrm{c}$$ है। उसके दोलन की आवृत्ति होगी :

A
$$\frac{1}{2 \pi \mathrm{t}_{\mathrm{0}}} \cos ^{-1}\left(\frac{\mathrm{a}+\mathrm{c}}{2 \mathrm{~b}}\right)$$
B
$$\frac{1}{2 \pi \mathrm{t}_{\mathrm{0}}} \cos ^{-1}\left(\frac{\mathrm{a}+\mathrm{b}}{2 \mathrm{c}}\right)$$
C
$$\frac{1}{2 \pi t_{\mathrm{0}}} \cos ^{-1}\left(\frac{2 \mathrm{a}+3 \mathrm{c}}{\mathrm{b}}\right)$$
D
$$\frac{1}{2 \pi t_{0}} \cos ^{-1}\left(\frac{a+2 b}{3 c}\right)$$

Questions Asked from Simple Harmonic Motion

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