JEE Main SHM PYQ

11.(JEE Main 2021 (Online) 22th July Evening Shift)

T₀ is the time period of a simple pendulum at a place. if the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, the modified time period is :

A. 4 T₀

B. $\frac{1}{4}$ T₀

C. T₀

D. 8 $π$ T₀

Correct Answer is Option (B)

$T = 2 π \sqrt{\frac{l}{g}}$

$T^{'} = \frac{T_{0}}{4}$

12.(JEE Main 2021 (Online) 18th March Morning Shift)

The time period of a simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ . The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-

A. 2%

B. 3%

C. 5%

D. 4%

Correct Answer is Option (B)

$T = 2 π \sqrt{\frac{l}{g}}$

$T^{2} = 2 π (\frac{l}{g})$

$g = 2 π \frac{l}{T^{2}}$

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

$\frac{Δ g}{g} = \frac{1 \times 10^{- 3}}{1 \times 10^{- 2}} + \frac{2 \times 1}{100}$

$\frac{Δ g}{g} = 0.02 + 0.01 = 0.03$

$100 \times \frac{Δ g}{g} = 0.03 \times 100 = 3 %$

$\frac{Δ g}{g} \times 100 = 3 %$

13. (JEE Main 2021 (Online) 16th March Morning Shift)

Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be :

A. $\sqrt{3} T$

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

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

D. $\sqrt{\frac{3}{2}} T$

Correct Answer is Option (B)

When lift is stationary

$T = 2 π \sqrt{\frac{L}{g}}$

A pseudo force will act downwards when lift is moving upwards.

$∴$ $g_{e f f} = g + \frac{g}{2} = \frac{3 g}{2}$

$∴$ New time period

$T^{'} = 2 π \sqrt{\frac{L}{g_{e f f}}}$

$T^{'} = 2 π \sqrt{\frac{2 L}{3 g}}$

$∴$ $T^{'} = \sqrt{\frac{2}{3}} T$

14.(JEE Main 2021 (Online) 26th February Evening Shift)

Given below are two statements :

Statement I : A second's pendulum has a time period of 1 second.

Statement II : It takes precisely one second to move between the two extreme positions.

In the light of the above statements, choose the correct answer from the options given below :

A. Both Statement I and Statement II are false

B. Statement I is false but Statement II is true

C. Both Statement I and Statement II are true

D. Statement I is true but Statement II is false

Correct Answer is Option (B)

As we know time period of second’s penduklum is 2 sec, so statement (1) is incorrect.

Time taken by particle performing SHM between two extreme position is half of the time period.

Here, T = 2 sec.

So, time = 2/2 = 1

15.(JEE Main 2021 (Online) 25th February Morning Shift)

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :

A. 16 m/s²

B. 2 $π$ ² ms^{$-$ 2}

C. $π$ ² ms^{$-$ 2}

D. 9.8 ms^{$-$ 2}

Correct Answer is Option (B)

The formula for the period of a simple pendulum is given by:

$T = 2 π \sqrt{\frac{l}{g}}$

where:

T is the period of the pendulum,
l is the length of the pendulum, and
g is the acceleration due to gravity.

We need to find g. Rearranging the formula for g, we get:

$g = \frac{4 π^{2} l}{T^{2}}$

Given:

l = 2 m,
T = 2 s,

Substituting these values into the equation, we get:

$g = \frac{4 π^{2} \times 2}{(2)^{2}} = 2 π^{2} m s^{- 2}$

Simple Harmonic Motion