Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 175 N/m. The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.

To find the spring constant of spring 2, we can use the equation for the period of simple harmonic motion.

The period, T, of an object undergoing simple harmonic motion is given by:

T = 2π√(m/k)

where m is the mass of the object and k is the spring constant.

Given that the mass of the objects on both springs is equal and their maximum velocities are the same, we can assume that the periods of their oscillations are also equal.

Let's denote the spring constant of spring 2 as k2.

From the information given, we know that the amplitude, A2, of the object on spring 2 is half that of the object on spring 1. The amplitude is related to the spring constant by the equation:

A = √(m/k)

Since the amplitude is proportional to the square root of the spring constant, we can write the following relationship:

A2 / A1 = √(k1 / k2)

where A1 is the amplitude of the object on spring 1.

Given that A2 = A1 / 2, we can substitute these values into the equation:

(A1 / 2) / A1 = √(k1 / k2)

Simplifying this equation, we get:

1/2 = √(k1 / k2)

Now, let's solve for k2:

Square both sides of the equation:

(1/2)^2 = ( √(k1 / k2) )^2

1/4 = k1 / k2

Rearranging the equation, we get:

k2 = k1 / (1/4)

Dividing k1 by 1/4 is equivalent to multiplying k1 by 4, so we have:

k2 = 4 * k1

Finally, substituting the value of k1 (175 N/m), we can find the spring constant of spring 2:

k2 = 4 * 175 N/m
k2 = 700 N/m

Therefore, the spring constant of spring 2 is 700 N/m.