JEE Main SHM PYQ

16. (JEE Main 2021 (Online) 24th February Evening Shift)

The period of oscillation of a simple pendulum is $T = 2 π \sqrt{\frac{L}{g}}$ . Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :

A. 1.30%

B. 1.33%

C. 1.13%

D. 1.03%

Correct Answer is Option (C)

Given, $T = 2 π \sqrt{\frac{L}{g}}$ .... (i)

where, time period, T = 1.95 s

Length of string, l = 1 m

Acceleration due to gravity = g

Error in time period, $Δ$ T = 0.01 s = 10^{$-$ 2} s

Error in length, $Δ$ L = 1 mm = 1 $\times$ 10^{$-$ 3} m

Squaring Eq. (i) on both sides, we get

$T^{2} = 4 π^{2} \frac{L}{g}$

$\Rightarrow g = 4 π^{2} \frac{L}{T^{2}}$

$\Rightarrow \frac{Δ g}{g} = \frac{Δ L}{L} + \frac{2 Δ T}{T} = \frac{10^{- 3}}{1} + \frac{2 \times 10^{- 2}}{1.95}$

$= 10^{- 3} + 1.025 \times 10^{- 2}$

$= 10^{- 3} + 10.25 \times 10^{- 3}$

$= 11.25 \times 10^{- 3}$

$∵$ $Δ g / g \times 100 = 11.25 \times 10^{- 3} \times 10^{2}$

$= 1.125 % ≃ 1.13 %$

17. (JEE Main 2021 (Online) 25th July Morning Shift )

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of 60 $^{\circ}$ to the vertical will be (g = 10 m/s²) ____________ m/s.

Correct Answer is (2)

JEE Main 2021 (Online) 25th July Morning Shift Physics - Simple Harmonic Motion Question 52 English Explanation

Applying work energy theorem :

w_g + w_T = $Δ$ K

$-$ mgl(1 $-$ cos60 $^{\circ}$ ) = $\frac{1}{2}$ mv² $-$ $\frac{1}{2}$ mu²

v² = u² $-$ 2gl(1 $-$ cos60 $^{\circ}$ )

v² = 9 $-$ 2 $\times$ 10 $\times$ 0.5 $(\frac{1}{2})$

v² = 4

v = 2 m/s

18. (JEE Main 2020 (Online) 8th January Evening Slot)

A simple pendulum is being used to determine th value of gravitational acceleration g at a certain place. Th length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :

A. 4.40%

B. 3.40%

C. 2.40%

D. 5.40%

Correct Answer is Option (A)

T = $2 π \sqrt{\frac{l}{g}}$

$\Rightarrow$ $g = \frac{4 π^{2} l}{T^{2}}$

$\Rightarrow$ $\frac{Δ g}{g} = \frac{Δ l}{l} + 2 \frac{Δ T}{T}$

= $\frac{0.1}{25} + 2 \frac{1}{50}$ = $\frac{11}{250}$

$∴$ % accuracy = $\frac{11}{250}$ $\times$ 100% = 4.40 %

19. (JEE Main 2019 (Online) 11th January Evening Slot)

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10^–2 m. The relative change in the angular frequency of the pendulum is best given by :

A. 1 rad/s

B. 10^{$-$ 3} rad/s

C. 10^{$-$ 1} rad/s

D. 10^{$-$ 5} rad/s

Correct Answer is Option (B)

Angular frequency of pendulum

$ω$ = $\sqrt{\frac{g_{e f f}}{ℓ}}$

$∴$ $\frac{Δ ω}{ω}$ = $\frac{1}{2}$ $\frac{Δ g_{e f f}}{g_{e f f}}$

$Δ$ $ω$ = $\frac{1}{2}$ $\frac{Δ g}{g} \times ω$

[ $ω_{s}$ = angular frequency of support]

$Δ$ $ω$ = $\frac{1}{2} \times \frac{2 A ω_{s}^{2}}{100} \times 100$

$Δ ω = 10^{- 3}$ rad/sec.

20. (JEE Main 2019 (Online) 11th January Evening Slot)

A pendulum is executing simple harmonic motion and its maximum kinetic energy is K₁. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K₂. Then :

A. $K_{2}$ = $\frac{K_{1}}{2}$

B. K₂ = 2K₁

C. K₂ = K₁

D. K₂ = $\frac{K_{1}}{4}$

Correct Answer is Option (B)

Maximum kinetic energy at lowest point B is given by

K = mg $ℓ$ (1 $-$ cos $θ$ )

where $θ$ = angular amp.

JEE Main 2019 (Online) 11th January Evening Slot Physics - Simple Harmonic Motion Question 89 English Explanation
K₁ = mg $ℓ$ (1 $-$ cos $θ$ )

K₂ = mg(2 $ℓ$ ) (1 $-$ cos $θ$ )

K₂ = 2K₁.