JEE Main SHM PYQ

6. ( JEE Main 2022 (Online) 28th July Evening Shift)

Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is -

(consider radius of earth $R_{E} = 6400 km$ and $g$ on earth $10 m / s^{2}$ )

A. 1200 km

B. 1600 km

C. 3200 km

D. 4800 km

Correct Answer is Option (C)

$T \propto \sqrt{1 / g}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{g_{2}}{g_{1}}} = \frac{R}{R + h}$

$\frac{4}{6} = \frac{R}{R + h}$

$\Rightarrow h = R / 2$

$= 3200$ km

7. (JEE Main 2022 (Online) 29th June Evening Shift)

The motion of a simple pendulum executing S.H.M. is represented by the following equation.

$y = A \sin (π t + ϕ)$ , where time is measured in second. The length of pendulum is

A. 97.23 cm

B. 25.3 cm

C. 99.4 cm

D. 406.1 cm

Correct Answer is Option (C)

$ω = π = \sqrt{\frac{g}{l}}$

So, $l = \frac{g}{π^{2}}$

$≃ 99.4$ cm (Nearest value)

8.(JEE Main 2022 (Online) 26th June Morning Shift)

Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with $\frac{g}{6}$ vertically upwards then the time period will be :

(Where g = acceleration due to gravity)

A. $\sqrt{\frac{6}{5}} T$

B. $\sqrt{\frac{5}{6}} T$

C. $\sqrt{\frac{5}{6}} T$

D. $\sqrt{\frac{7}{6}} T$

Correct Answer is Option (C)

$T^{'} = 2 π \sqrt{\frac{I}{g_{eff}}}$

$T^{'} = 2 π \sqrt{\frac{I}{g + \frac{g}{6}}} = 2 π \sqrt{\frac{6 l}{7 g}}$

$\Rightarrow T^{'} = \sqrt{\frac{6}{7}} T$

9. (JEE Main 2022 (Online) 28th June Morning Shift )

A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60 $^{\circ}$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms^{$-$ 1}. (if g = 10 m/s²)

JEE Main 2022 (Online) 28th June Morning Shift Physics - Simple Harmonic Motion Question 37 English

Correct Answer is (5)

$\frac{1}{2} m v^{2} = m g l (1 - \cos θ)$

$\Rightarrow v = \sqrt{2 g l (1 - \cos θ)}$

$= \sqrt{2 \times 10 \times 2.5 \times \frac{1}{2}}$

$= 5$ m/s

10. (JEE Main 2021 (Online) 31st August Evening Shift)

A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $\frac{1}{4}$ times that of the bob and the length of the thread is increased by 1/3^rd of the original length, then the time period of the simple harmonic oscillations will be :-

A. T

B. $\frac{3}{2}$ T

C. $\frac{3}{4}$ T

D. $\frac{4}{3}$ T

Correct Answer is Option (D)

$T = 2 π \sqrt{l / g}$

When bob is immersed in liquid

mg_eff = mg $-$ Buoyant force

mg_eff = mg $-$ v $σ$ g ( $σ$ = density of liquid)

$= m g - v \frac{ρ}{4} g$

$= m g - \frac{m g}{4} = \frac{3 m g}{4}$

$∴$ $g_{e f f} = \frac{3 g}{4}$

$T_{1} = 2 π \sqrt{\frac{l_{1}}{g_{e f f}}}$

$l_{1} = l + \frac{l}{3} = \frac{4 l}{3}, l_{e f f} = \frac{3 g}{4}$

By solving

$T_{1} = \frac{4}{3} 2 π \sqrt{l / g}$

$T_{1} = \frac{4 T}{3}$