Question

Груз массой 0,3 кг подвешенный на пружине совершает гармонические колебания. К нему прикрепляют груз массой 0,9 кг. Во сколько раз увеличиться период колебаний?

Answer 1

T (1)=2π√m/k
k- жесткость пружины, она не меняется
T (2)=2π√M/k
M=m+0,9
Из отношения T (2)/T (1)получим:
T (2)/T (1)=√M/m=√1,2/0,3= 2
период увеличился в 2 раза
 

Answer 2

Harmonic Oscillation and Period Change

When a mass of 0.3 kg is suspended on a spring and undergoes harmonic oscillations, the period of oscillation can be calculated using the formula:

T = 2π√(m/k)

Where: - T = period of oscillation - m = mass - k = spring constant

Now, when an additional mass of 0.9 kg is attached to the original mass, the total mass becomes 1.2 kg. The period of oscillation for this new system can be calculated using the same formula.

To find the ratio of the new period to the original period, we can use the following relationship:

T₂ / T₁ = √(m₂ / m₁)

Where: - T₂ = period of oscillation with the added mass - T₁ = original period of oscillation - m₂ = total mass after adding the additional mass - m₁ = original mass

Let's calculate the ratio of the new period to the original period using the given masses.

The original mass (m₁) is 0.3 kg, and the additional mass (m₂) is 0.9 kg, making the total mass 1.2 kg.

Using the formula, we get:

T₂ / T₁ = √(1.2 / 0.3) = √4 = 2

So, the period of oscillation will increase by a factor of 2 when the additional mass is attached to the original mass.

This means that the period of oscillation will double when the additional mass is added to the system.