3springs mass systems in vertical plane .the mass of all pulleys and connecting strings and springs are negligible and friction at all contact is absent .calculate the T1,t2,T3 WHICH ARE TIME PERIODS OF SMALL VERTICAL OSCILLATION OF MASS IN SYSTEM 1,2,3

1.to the horizontal bar one string and spring of4k are attached connected to pulleys one pulley is attached to the bottom with spring of 8k,other pulley suspended to mass m

2.from the horizontal bar 2 springs 8k and 4k are attached.8k spring is connected to pulley which is attached to a string to the bottom other pulley attached to spring 4k has mass m suspended to it

3. from the horizontal bar one spring and a string are attached spring is of 4k.it is attached to pulley which is connected to another spring of 8k attached to the bottom,other pulley is suspended with mass m it is attached to the top with a string

To analyze the three spring-mass systems you've described, we need to determine the time periods of small vertical oscillations for each system. The time period \( T \) of a mass-spring system can be calculated using the formula:

Time Period Formula

The formula for the time period \( T \) of a simple harmonic oscillator is given by:

T = 2π√(m/k)

where:

• T = time period
• m = mass attached to the spring
• k = spring constant

System 1 Analysis

In the first system, we have a mass \( m \) suspended from a pulley, with two springs connected: one with a spring constant of \( 4k \) and another with \( 8k \). The effective spring constant \( k_{eff} \) for springs in series is given by:

1/k_{eff} = 1/k_1 + 1/k_2

Substituting the values:

1/k_{eff} = 1/(4k) + 1/(8k) = 2/(8k) + 1/(8k) = 3/(8k)

Thus, the effective spring constant is:

k_{eff} = 8k/3

Now, using the time period formula:

T_1 = 2π√(m/(8k/3)) = 2π√(3m/(8k))

System 2 Analysis

For the second system, we have two springs: one with \( 8k \) connected to a pulley and another with \( 4k \) connected to a mass \( m \). The effective spring constant for these two springs in series is:

1/k_{eff} = 1/(8k) + 1/(4k) = 1/(8k) + 2/(8k) = 3/(8k)

Thus, the effective spring constant is:

k_{eff} = 8k/3

Using the time period formula again:

T_2 = 2π√(m/(8k/3)) = 2π√(3m/(8k))

System 3 Analysis

In the third system, we have a spring of \( 4k \) attached to a pulley, which is then connected to another spring of \( 8k \) at the bottom. The mass \( m \) is suspended from the pulley. The effective spring constant for these two springs in series is:

1/k_{eff} = 1/(4k) + 1/(8k) = 2/(8k) + 1/(8k) = 3/(8k)

Thus, the effective spring constant is:

k_{eff} = 8k/3

Using the time period formula:

T_3 = 2π√(m/(8k/3)) = 2π√(3m/(8k))

Summary of Time Periods

In summary, the time periods for the small vertical oscillations of the masses in the three systems are:

• T_1 = 2π√(3m/(8k))
• T_2 = 2π√(3m/(8k))
• T_3 = 2π√(3m/(8k))

Interestingly, all three systems yield the same time period due to the effective spring constants being identical. This illustrates how different configurations can lead to similar dynamic behaviors in oscillatory systems.