Simple Harmonic Motion JEE Advanced Previous Year Questions With Solutions

Simple Harmonic Motion or SHM is defined as a motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards the mean position. The acceleration of a particle executing simple harmonic motion is given by, a(t) = -ω² x(t). Here, ω is the angular velocity of the particle.

Download Simple Harmonic Motion JEE Advanced Previous Year Questions With Solutions PDF

Question 1) A particle undergoing simple harmonic motion has time-dependent displacement given by x(t) = A sin (πt/90). The ratio of kinetic to the potential energy of this particle at t = 210 s will be

(A) 1/9

(B) 1

(C) 2

(D) ⅓

Answer: (D) ⅓

Solution:

Kinetic energy, k = (½)mω²A²cos²ωt

Potential energy, U = (½)mω²A²sin²ωt

(k/U) = cot²ωt =cot²(π/90)(210) = 1/3

Question 2) A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential energy (PE) equals kinetic energy (KE), the position of the particle will be

(A) A/2

(B) A/2√2

(C) A/√2

(D) A

Answer: (C) A/√2

Solution:

Potential energy (U) = (½)kx²

Kinetic energy (K) = (½)kA² – (½)kx²

Given, U = K

(½)kx²= (½)kA² – (½)kx²

\(\begin{array}{l}x = \pm \frac{A}{\sqrt{2}}\end{array} \)

Question 3) A simple harmonic motion is represented by y = 5(sin 3πt + √3 cos 3πt ) cm. The amplitude and time period of the motion is

(A) 10 cm, (⅔) s

(B) 10 cm, (3/2) s

(C) 5 cm, (3/2) s

(D) 5 cm, (⅔) s

Answer: (A) 10 cm, (⅔) s

Solution:

Given, y = 5(sin 3πt + √3 cos 3πt )

⇒y = 10 sin (3πt +π/3)

Therefore, Amplitude = 10 cm

Time period, T = 2π/ω = 2π/3π = (⅔) s

Question 4) The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be:

(A) (√3/2)s

(B) (2/√3)s

(C) (3/2)s

(D) 2√3s

Answer: (D) 2√3s

Solution:

Acceleration due to gravity g = GM/R²

\(\begin{array}{l}\frac{g_{p}}{g_{e}}=\frac{M_{p}}{M_{e}}(\frac{R_{e}}{R_{p}})^{2}=3(1/3)^{2}=1/3\end{array} \)

Also, T ∝ 1/√g

\(\begin{array}{l}\Rightarrow \frac{T_{p}}{T_{e}}=\sqrt{\frac{g_{p}}{g_{e}}}=\sqrt{3}\end{array} \)

⇒ T_p = 2√3 s

Question 5) A particle executes simple harmonic motion with a frequency f. The frequency with which kinetic energy oscillates is

(A) f/2

(B) f

(C) 2f

(D) 4f

Answer: (C) 2f

Solution:

During one complete oscillation, the kinetic energy of a particle executing SHM will become maximum twice. Therefore, the frequency with which its kinetic energy oscillates will be 2f.

Question 6) A object of mass 0.2 kg executes a simple harmonic oscillation along the x-axis with a frequency of (25/π) Hz. At position x = 0.04, the object has a kinetic energy of 0.5 J and potential energy 0.4 J. The amplitude of oscillation is _____m

Answer: 0.06 m

Solution:

Given: m = 0.2 kg, ω = 25/π, K.E = 0.5 J, P.E = 0.4 J

T.E = K.E + P.E

= 0.5 + 0.4 = 0.9 J

Also, T.E = (½)mω²a² = (½)m x 4π²v²a²

⇒ 0.9 = (½) x 0.2 x 4π² x (25/π) x (25/π) x a²

Therefore, a = 0.06 m

Question 7) A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10¹²/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. Of silver = 108 and Avogadro’s number = 6.02 x 10²³ gm/mole)

(A) 6.4 N/m

(B) 7.1 N/m

(C) 2.2 N/m

(D) 5.5 N/m

Answer: (B) 7.1 N/m

Solution:

As we know, frequency in SHM

f = (1/2π)(√k/m) = 10¹²

Where m = mass of one atom

Mass of one atom of silver = [108/(6.02 x 10²³)] x 10^-3 kg

\(\begin{array}{l}f = \frac{1}{2\pi }\sqrt{\frac{k}{108\times 10^{-3}}\times 6.02 \times 10^{23}}=10^{12}\end{array} \)

Solving we get, spring constant

K = 7.1 N/m

Question 8) A particle moves with simple harmonic motion in a straight line. In the first τ s, after starting from rest, it travels a distance a, and in the next τ s it travels 2a, in the same direction, then:

(A) Amplitude of motion in 3a

(B) the time period of oscillations is 8τ

(C) Amplitude of motion in 4a

(D) the time period of oscillations is 6τ

Answer: (D) time period of oscillations is 6τ

Solution:

Question 9) A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 N/m and oscillates in a medium of constant 10^-2 kg/s. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to

(A) 2 s

(B) 3.5 s

(C) 5 s

(D) 7 s

Answer: (B) 3.5 s

Solution:

Since the system dissipates its energy gradually, and hence amplitude will also decrease with time according to

a = a₀e^bt/m——————(1)

Energy of vibration drop to half of its initial value (E₀), as E ∝ a² ⇒ a ∝ √E

a = a₀/√2

⇒ (bt/m) = 10^-2t/0.1 = t/10

From equation (1)

(a₀/√2) = a₀e^-t/10

1/√2 = e^-t/10

√2 = e^t/10

In √2 = t/10

t = 3.5 seconds

Question 10) A simple pendulum oscillating in the air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is (1/16)th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is

(A) 2T√1/10

(B) 2T√1/14

(C) 4T√1/15

(D) 4T√1/14

Answer: (C) 4T√1/15

Solution:

T = 2π√l/g_eff

When immersed in non-viscous liquid

g_eff= (g – (g/16)) = 15g/16

Now

\(\begin{array}{l}\frac{T’}{T}=\frac{2\pi \sqrt{\frac{L}{15g/16}}}{2\pi \sqrt{\frac{L}{g}}}\end{array} \)

\(\begin{array}{l}\Rightarrow T’ = \frac{4T}{\sqrt{15}}\end{array} \)

Simple Harmonic Motion (SHM) – JEE Concept

Also Read:

Simple Harmonic Motion JEE Main Previous Year Questions With Solutions

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