JEE Main SHM PYQ

16. (JEE Main 2021 (Online) 31st August Morning Shift )

A particle of mass 1 kg is hanging from a spring of force constant 100 Nm^{$-$ 1}. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is $\frac{T}{x}$ . The value of x is _____________.

Correct Answer is (8)

JEE Main 2021 (Online) 31st August Morning Shift Physics - Simple Harmonic Motion Question 44 English Explanation 1
KE = PE

$y = \frac{A}{\sqrt{2}} = A \sin ω t$

$t = \frac{T}{8} = \frac{T}{x}$

x = 8

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

In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be ____________ rad/s when k = 20 N/m.

JEE Main 2021 (Online) 25th July Morning Shift Physics - Simple Harmonic Motion Question 53 English

Correct Answer is (10)

$ω = \sqrt{\frac{k_{e q}}{μ}}$

$μ$ = reduced mass

springs are in series connection

$k_{e q} = \frac{k_{1} k_{2}}{k_{1} + k_{2}}$

$k_{e q} = \frac{k \times 4 k}{5 k} = \frac{4 k}{5}$

$k_{e q} = \frac{4 \times 20}{5}$ N/m = 16 N/m

$μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{0.2 \times 0.8}{0.2 + 0.8} = 0.16$ kg

$ω = \sqrt{\frac{16}{0.16}} = \sqrt{100} = 10$

18 .(JEE Main 2021 (Online) 17th March Morning Shift )

Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is T_b/T_a = $\sqrt{x}$ , where value of x is ___________. (Round off to the Nearest Integer)

JEE Main 2021 (Online) 17th March Morning Shift Physics - Simple Harmonic Motion Question 62 English

Correct Answer is (2)

$T_{a} = 2 π \sqrt{\frac{M}{k}}$

$K_{e q_{s e r i e s}} = \frac{k_{1} k_{2}}{k_{1} + k_{2}} = \frac{k}{2}$

$T_{b} = 2 π \sqrt{\frac{M}{k / 2}} = 2 π \sqrt{\frac{2 M}{k}}$

$T_{b} = \sqrt{2} T_{a}$

$\frac{T_{b}}{T_{a}} = \sqrt{2}$

$∴$ x = 2

19.(JEE Main 2020 (Online) 6th September Evening Slot)

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by
y(t) = y₀ sin² $ω$ t, where 'y' is measured from the lower end of unstretched spring. Then $ω$ is:

A. $\sqrt{\frac{g}{y_{0}}}$

B. $\frac{1}{2} \sqrt{\frac{g}{y_{0}}}$

C. $\sqrt{\frac{2 g}{y_{0}}}$

D. $\sqrt{\frac{g}{2 y_{0}}}$

Correct Answer is Option (D)

JEE Main 2020 (Online) 6th September Evening Slot Physics - Simple Harmonic Motion Question 79 English Explanation
y(t) = y₀ sin² $ω$ t

= $\frac{1}{2} y_{0} (2 \sin^{2} ω t)$

= $\frac{1}{2} y_{0} (1 - \cos 2 ω t)$

From comparing standard equation of SHM Amplitude A = $\frac{y_{0}}{2}$

And frequency = 2 $ω$

At equilibrium situation, $\frac{m g}{k} = \frac{y_{0}}{2}$

$\Rightarrow$ $\frac{k}{m} = \frac{2 g}{y_{0}}$

$∴$ 2 $ω$ = $\sqrt{\frac{k}{m}}$

$\Rightarrow$ 2 $ω$ = $\sqrt{\frac{2 g}{y_{0}}}$

$\Rightarrow$ $ω$ = $\sqrt{\frac{2 g}{4 y_{0}}}$ = $\sqrt{\frac{g}{2 y_{0}}}$

20.(JEE Main 2020 (Online) 3rd September Evening Slot)

A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of f is :

A. 1

B. $\frac{1}{2}$

C. $\sqrt{2}$

D. $\frac{1}{\sqrt{2}}$

Correct Answer is Option (D)

At equilibrium position

V₀ = V

$V_{0} = ω_{1} A = \sqrt{\frac{K}{m}} A$ .....(i)

$V = ω A^{1} = \sqrt{\frac{K}{\frac{m}{2}}} A^{1}$ .....(ii)

$∴$ $A^{1} = \frac{A}{\sqrt{2}}$