A block with mass 0.55 kg on a frictionless surface is attached to a spring with spring constant 39 N/m. The block is pulled from the equilibrium position and released. What is the period of the system?

answer in seconds...

To find the period of the system, we can use the formula:

T = 2π * √(m / k)

where:
T is the period of the system
π is a constant approximately equal to 3.14159
m is the mass of the block (0.55 kg in this case)
k is the spring constant (39 N/m in this case)

Let's substitute the given values into the formula and calculate the period.

T = 2π * √(0.55 kg / 39 N/m)

T = 2π * √(0.0141 kg/m)

Now, let's calculate the square root part of the equation.

√(0.0141 kg/m) ≈ 0.1189 s

Finally, substitute the value back into the equation:

T = 2π * 0.1189 s

T ≈ 0.747 s

Therefore, the period of the system is approximately 0.747 seconds.

To find the period of the system, we need to use the equation for the period of a mass-spring system. The period (T) is given by the formula:

T = 2π√(m/k)

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

In this case, the mass (m) is 0.55 kg and the spring constant (k) is 39 N/m. Plugging these values into the formula, we get:

T = 2π√(0.55/39)

Calculating this expression, we find the value of T to be approximately 0.359 seconds. Therefore, the period of the system is 0.359 seconds.