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A mass loaded spring is displaced 5 cm below its equilibrium position and then released, it travels from the lowest point to the highest point within 0.25 sec. Determine, the maximum time required for the system to oscillate from 5cm below the equilibrium position to 3cm above equilibrium position.​

User Foglerit
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Answer:

The maximum time required for the system to oscillate from 5 cm below the equilibrium position to 3 cm above the equilibrium position is approximately 1.309 seconds.

Step-by-step explanation:

The time period (T) of a mass-spring system is given by:

T = 2π√(m/k)

where m is the mass attached to the spring, and k is the spring constant.

Given that the spring is displaced 5 cm below its equilibrium position and travels from the lowest point to the highest point within 0.25 sec. This means that the time period of the system is:

T = 2(0.25) = 0.5 sec

Now, let's assume that the maximum time required for the system to oscillate from 5 cm below the equilibrium position to 3 cm above the equilibrium position is t seconds.

So, the time taken for the system to move from the lowest point to 3 cm above the equilibrium position is (t/2) seconds.

According to the given problem, the displacement is 5 cm below the equilibrium position, so the amplitude of oscillation is:

A = (5 + 3) / 2 = 4 cm

Now, using the formula for time period, we get:

T = 2π√(m/k) ---- (1)

We know that the maximum displacement (amplitude) of oscillation, A = 4 cm. This can be expressed in terms of mass and spring constant as:

A = (m * g) / k ---- (2)

where g is the acceleration due to gravity.

Squaring equation (2) and solving for m/k, we get:

(m/k) = (A * k) / g)^2 ---- (3)

Substituting equation (3) into equation (1), we get:

T = 2π√[((A * k) / g)^2] ---- (4)

Simplifying equation (4), we get:

T = 2π * (A / g) * √(1/k) ---- (5)

Now, substituting the values of T, A, and g into equation (5), we get:

0.5 = 2π * (4 / 9.8) * √(1/k)

Simplifying this equation, we get:

√(k) = 2π * (4 / 9.8) / 0.5

√(k) = 10.239

k = 105

So, the spring constant is 105 N/m.

Now, substituting the value of k into equation (3), we get:

(m/k) = (A * k / g)^2

(m/k) = (4 * 105 / 9.8)^2

(m/k) = 73.88

So, the mass attached to the spring is:

m = (73.88) * (105)

m = 7757.4 g

m = 7.7574 kg

Now, we know the mass of the system and the spring constant, we can calculate the maximum time required for the system to oscillate from 5 cm below the equilibrium position to 3 cm above the equilibrium position.

The time period (T) of the system is given by:

T = 2π√(m/k)

T = 2π√(7.7574/105)

T = 1.309 sec (approx)

Therefore, the maximum time required for the system to oscillate from 5 cm below the equilibrium position to 3 cm above the equilibrium position is approximately 1.309 seconds.

User Capagris
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