When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12 Hz. When another object of mass m2 is hung on the spring along with m1, the frequency of the motion is 4 Hz. Find the ratio m2/m1 of the masses.

w= sqrt (k/m)

w^2= k/m

(w1/w2)^2=9 = (m1+m2)/m1

9= 1 + m2/m1

check my thinking.

I'm sorry but can you please clarify your response. I am not clear on your explanation.

A body is supported by a spiral spring and causes a stretch of 1.5cm in the spring. If the mass is now set in vertical oscillation of small amplitude, what is the periodic time of oscillation

A body is supported by a spiral spring and causes a stretch of 1.5cm in the spring. If the mass is now set in vertical oscillation of small amplitude, what is the periodic time of oscillation

To determine the periodic time of oscillation, we need to use the formula:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

Since the problem states that the body causes a stretch of 1.5 cm in the spring, we can assume that the displacement is equal to the amplitude of oscillation.

Let's assume the spring constant of the spiral spring is denoted by k.

Therefore, k = F/x

where F is the force applied to stretch the spring and x is the displacement.

Without knowing the exact values of F and x, we won't be able to find the value of k and determine the periodic time of oscillation.

To find the period of oscillation, we need to determine the angular frequency (ω) and then use it to calculate the period (T).

The angular frequency (ω) can be found using the formula:

ω = √(k/m)

Where:
k is the spring constant (also known as the force constant), and
m is the mass of the object.

To find the spring constant (k), we can use Hooke's Law, which states that the force applied by a spring is directly proportional to the displacement (stretch or compression) of the spring. Mathematically, it can be written as:

F = -kx

Where:
F is the force applied by the spring,
k is the spring constant, and
x is the displacement of the spring.

In this case, the displacement of the spring is given as 1.5 cm. Let's convert it to meters by dividing by 100:

x = 1.5/100 = 0.015 m

The force (F) is equal to the weight of the object, which can be calculated as:

F = mg

Where:
m is the mass of the object, and
g is the acceleration due to gravity, approximately 9.8 m/s^2.

We need to rearrange the equation to solve for k:

k = -F/x = -mg/x

Substituting the given values:

k = -(m * 9.8) / 0.015

Now, we have the value of the spring constant (k). Let's substitute it back into the formula for ω:

ω = √(k/m) = √(-(m * 9.8) / 0.015m) = √(-9.8 / 0.015)

Now that we have ω, we can calculate the period (T) using the formula:

T = 2π / ω

Substitute the value of ω:

T = 2π / √(-9.8 / 0.015)

Calculate the square root and divide 2π by the result to obtain the value of T.

frequency f = ω/2π = sqrt(k/m)/2π since k is a constant for the spring.f varies as 1/sqrt(m)

f*sqrt(m) = constant
or
12*sqrt(m1) = 4*sqrt(m1+m2)

simplifying 144*m1 = 16*(m1+m2)

or 144/16 = (m1 + m2)/m1

9 = 1+m2/m1

or m2/m1 = 9 - 1 = 8