JEE Main SHM PYQ

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

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :

A. $\frac{\sqrt{3}}{2}$ s

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

C. $\frac{2}{\sqrt{3}}$ s

D. $2 \sqrt{3}$ s

Correct Answer is Option (D)

$∵$     g = $\frac{G M}{R^{2}}$

$\frac{g_{p}}{g_{e}}$ = $\frac{M_{e}}{M_{e}} {(\frac{R_{e}}{R_{p}})}^{2}$ = 3 ${(\frac{1}{3})}^{2}$ = $\frac{1}{3}$

Also T $\propto$ $\frac{1}{\sqrt{g}}$

$\Rightarrow$    $\frac{T_{p}}{T_{e}}$ = $\sqrt{\frac{g_{e}}{g_{p}}}$ = $\sqrt{3}$

$\Rightarrow$   T_p = 2 $\sqrt{3}$ s

22. (JEE Main 2015 (Offline))

The period of oscillation of a simple pendulum is $T = 2 π \sqrt{\frac{L}{g}}$ . Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

A. 1 %

B. 5 %

C. 2 %

D. 3 %

Correct Answer is Option (D)

Given $T = 2 π \sqrt{\frac{L}{g}}$

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

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

[ as $T = \frac{t}{n}$ ]

So, percentage error in $g$ =

$\frac{Δ g}{g} \times 100 = \frac{Δ L}{L} \times 100 + 2 \frac{Δ t}{t} \times 100$

= $\frac{0.1}{20.0} \times 100 + 2 \times \frac{1}{90} \times 100$

= 2.72 % = 3 %

23. (JEE Main 2015 (Offline))

A pendulum made of a uniform wire of cross sectional area $A$ has time period $T .$ When an additional mass $M$ is added to its bob, the time period changes to $T_{M .}$ If the Young's modulus of the material of the wire is $Y$ then $\frac{1}{Y}$ is equal to :
( $g =$ $g r a v i t a t i o n a l$ $a c c e l e r a t i o n$ )

A. $[1 - {(\frac{T_{M}}{T})}^{2}] \frac{A}{M g}$

B. $[1 - {(\frac{T}{T_{M}})}^{2}] \frac{A}{M g}$

C. $[{(\frac{T_{M}}{T})}^{2} - 1] \frac{A}{M g}$

D. $[{(\frac{T_{M}}{T})}^{2} - 1] \frac{M g}{A}$

Correct Answer is Option (C)

As we know, time period, $T = 2 π \sqrt{\frac{ℓ}{g}}$

When a additional mass $M$ is added then

$T_{M} = 2 π \sqrt{\frac{ℓ + Δ ℓ}{g}}$

$\frac{T_{M}}{T} = \sqrt{\frac{ℓ + Δ ℓ}{ℓ}}$

or, ${(\frac{T_{M}}{T})}^{2} = \frac{ℓ + Δ ℓ}{ℓ}$

or, ${(\frac{T_{M}}{T})}^{2} = 1 + \frac{M g}{A y}$

$[$ as $Δ ℓ = \frac{M g ℓ}{A y}]$

$∴$ $\frac{1}{y} = [{(\frac{T_{M}}{T})}^{2} - 1] \frac{A}{M g}$

24. (AIEEE 2005)

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

A. first decrease and then increase to the original value

B. first increase and then decrease to the original value

C. increase towards a saturation value

D. remain unchanged

Correct Answer is Option (B)

Center of mass of combination of liquid and hollow portion (at position $ℓ$ ), first goes down (to $ℓ + Δ ℓ$ ) and when total water is drained out, center of mass regain its original position (to $ℓ$ ), $∴$ $^{'} T^{'}$ first increases and then decreases to original value.

AIEEE 2005 Physics - Simple Harmonic Motion Question 124 English Explanation

25. (AIEEE 2004)

The bob of a simple pendulum executes simple harmonic motion in water with a period $t,$ while the period of oscillation of the bob is $t_{0}$ in air. Neglecting frictional force of water and given that the density of the bob is $(4 / 3) \times 1000 k g / m^{3} .$ What relationship between $t$ and $t_{0}$ is true

A. $t = 2 t_{0}$

B. $t = t_{0} / 2$

C. $t = t_{0}$

D. $t = 4 t_{0}$

Correct Answer is Option (A)

$t = 2 π \sqrt{\frac{ℓ}{g_{e f f}}}; t_{o} = 2 π \sqrt{\frac{ℓ}{g}}$

AIEEE 2004 Physics - Simple Harmonic Motion Question 131 English Explanation

$m g_{e f f} = m g - B = m y - V \times 100 \times g$

$∴$ $g_{e f f} = g - \frac{100}{(m / v)} g$

$= g - \frac{1000}{\frac{4}{3} \times 1000} g = \frac{g}{4}$

$∴$ $t = 2 π \sqrt{\frac{ℓ}{g / 4}} t = 2 t_{0}$