Two light springs of force constants k1 and k2, loaded with equal masses undergo vertical oscillations. If the maximum velocities of the masses are the same, the ratio of the amplitudes of oscillations is:a)k1/k2b)1c)√(k2/k1)d)none of theseCorrect answer is option 'C'. Can you explain this answer?

Question

Two light springs of force constants k1 and k2, loaded with equal masses undergo vertical oscillations. If the maximum velocities of the masses are the same, the ratio of the amplitudes of oscillations is:a)k1/k2b)1c)&#8730;(k2/k1)d)none of theseCorrect answer is option 'C'. Can you explain this answer?

Answer

As the maximum velocities are equal
a1ω1 = a2ω2
a1/a2 = ω2/ω1 ----- (1)
as, k1 = mω1² ----- (2)
k2 = mω2² ----- (3)
substituting eq 2 and eq 3 in eq 1 ,we get
a1/a2 = (k1/k2)^1/2

Answer

As the max. velocities are equal
a1ω1 = a2ω2
a1/a2 = ω2/ω1 ----- eq 1

as , k1 = mω1^2 ----- eq 2
k2 = mω2^2 ----- eq3
substituting eq 2 and eq 3 in eq 1 ,we get

a1/a2 = (k1/k2)^2

Answer

Solution:

Given:
Force constant of the first spring, k1
Force constant of the second spring, k2

Let's assume that the masses attached to both springs are m each.

Maximum velocity is the same for both masses.

When a mass is attached to a spring, it undergoes vertical oscillations. The equation of motion for the mass-spring system is given by:

m(d^2y/dt^2) + ky = 0

Where,
m = mass attached to the spring
y = displacement of the mass from its equilibrium position
k = force constant of the spring

We know that the force constant of a spring is given by k = mω^2, where ω is the angular frequency.

Therefore, the equation of motion can be written as:

(d^2y/dt^2) + (k/m)y = 0

For a simple harmonic motion, the angular frequency ω is given by:

ω = √(k/m)

The time period of oscillation is given by:

T = 2π/ω = 2π√(m/k)

From the given information, we know that the maximum velocities of the masses are the same. The maximum velocity of a mass in simple harmonic motion is given by:

vmax = ωA

Where,
vmax = maximum velocity
A = amplitude of oscillation

Since the maximum velocities of both masses are the same, we can write:

ω1A1 = ω2A2

Substituting the value of ω = √(k/m) in the above equation, we get:

√(k1/m)A1 = √(k2/m)A2

Canceling out the mass term and rearranging the equation, we get:

A1/A2 = √(k2/k1)

Therefore, the ratio of the amplitudes of oscillations is √(k2/k1), which is option C.

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