JEE Main SHM PYQ

1.(JEE Main 2021 (Online) 26th February Morning Shift)

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period :

A. $2 π \sqrt{\frac{R}{g}}$

B. $\frac{g}{2 π R}$

C. $\frac{2 π R}{g}$

D. $\frac{1}{2 π} \sqrt{\frac{g}{R}}$

Correct Answer is Option (A)

JEE Main 2021 (Online) 26th February Morning Shift Physics - Simple Harmonic Motion Question 70 English Explanation

Value of g on the particle of mass m,

$g = \frac{G M d}{R^{3}}$

Force acting on the particle towards the center of the earth,

F = mg

Force along the tunnel = F cos $θ$ = F₁

= mg cos $θ$

= m . $\frac{G M d}{R^{3}}$ $(\frac{x}{d})$

= $\frac{G M m}{R^{3}}$ . x

= $\frac{g_{s} m}{R} . x$ [as g_s = $\frac{G M}{R^{2}}$ at earth surface]

$∴$ acceleration along the tunnel

$α$ = $\frac{F_{1}}{m}$

= $\frac{g_{s}}{R} . x$

Time period = $\frac{2 π}{ω} = 2 π \sqrt{\frac{x}{α}}$ [as $ω^{2} = \frac{α}{x}$ ]

$= 2 π \sqrt{\frac{R}{g_{s}}}$

2. (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 $ω$ . If the radius of the bottle is 2.5 cm then $ω$ 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}

Correct Answer is Option (D)

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

F = $- (ρ A x) g$

$\Rightarrow$ ma = $- (ρ A x) g$

$\Rightarrow$ a = $- (\frac{ρ A g}{m}) x$

$∴$ $ω^{2} = \frac{ρ A g}{m}$ = $\frac{ρ (π r^{2}) g}{m}$

$\Rightarrow$ $ω$ = $\sqrt{\frac{10^{3} \times π \times {(2.5 \times 10^{- 2})}^{2} \times 10}{310 \times 10^{- 3}}}$ = 7.90 rad/s

3.(JEE Main 2018 (Offline))

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10¹²/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 10²³ gm mole^–1)

A. 5.5 N/m

B. 6.4 N/m

C. 7.1 N/m

D. 2.2 N/m

Correct Answer is Option (C)

6.02 $\times$ 10²³ atoms of silver = 108 gm

1 atoms of silver = $\frac{108 \times 10^{- 3}}{6.02 \times 10^{23}}$ kg

For a harmonic oscillator

f = $\frac{1}{2 π}$ $\sqrt{\frac{k}{m}}$

Where k = force constant

$\Rightarrow$ f² = $\frac{1}{4 π^{2}}$ $(\frac{k}{m})$

$\Rightarrow$ k = mf² $\times$ 4 $π$ ²

Given,

f = 10¹²

m = $\frac{108 \times 10^{- 3}}{6.02 \times 10^{23}}$

$∴$ k = $\frac{108 \times 10^{- 3}}{6.02 \times 10^{23}}$ $\times$ 10¹² $\times$ 4 $π$ ²

= 7.1 N/m

4.(JEE Main 2016 (Online) 10th April Morning Slot)

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

Correct Answer is Option (D)

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

$∴$    Maximum acceleration of the washer,

a_max = $ω$ ²A = g

$\Rightarrow$     $ω$ = $\sqrt{\frac{g}{A}}$ = $\sqrt{\frac{10}{0.07}}$ = $\sqrt{\frac{1000}{7}}$

$∴$    Frequency of the piston,

f = $\frac{ω}{2 π}$ = $\frac{1}{2 π} \sqrt{\frac{1000}{7}}$ = 1.9 Hz

5.(JEE Main 2013 (Offline))

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M .$ The piston and the cylinder have equal cross sectional area $A$ . When the piston is in equilibrium, the volume of the gas is $V_{0}$ and its pressure is $P_{0} .$ The piston is slightly displaced from the equilibrium position and released,. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frquency

A. $\frac{1}{2 π} \frac{A γ P_{0}}{V_{0} M}$

B. $\frac{1}{2 π} \frac{V_{0} M P_{0}}{A^{2} γ}$

C. $\frac{1}{2 π} \sqrt{\frac{A γ P_{0}}{V_{0} M}}$

D. $\frac{1}{2 π} \sqrt{\frac{M V_{0}}{A γ P_{0}}}$

Correct Answer is Option (C)

$\frac{M g}{A} = P_{0}$

$M g = P_{0} A . . . (1)$

$P_{0} V_{0}^{γ} = P V^{γ}$

$P = \frac{P_{0} x_{0}^{γ}}{{(x_{0} - x)}^{y}}$

Let piston is displaced by distance $x$

$M g - (\frac{P_{0} x_{0}^{γ}}{{(x_{0} - x)}^{γ}}) A = F_{r e s t o r i n g}$

JEE Main 2013 (Offline) Physics - Simple Harmonic Motion Question 111 English Explanation

$P_{0} A (1 - \frac{x_{0}^{γ}}{{(x_{0} - x)}^{γ}}) = F_{r e s t o r i n g}$

$[x_{0} - x \approx x_{0}]$

$F = - \frac{γ P_{0} A x}{x_{0}}$

$∴$ Frequency with which piston executes $S H M .$

$f = \frac{1}{2 π} \sqrt{\frac{γ P_{0} A}{x_{0} M}} = \frac{1}{2 π} \sqrt{\frac{γ P_{0} A^{2}}{M V_{0}}}$