A physics teacher designs a rather unusual demonstration of simple harmonic motion which utilizes a combination of a 4.0 kg

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a physics teacher designs a rather unusual demonstration of simple harmonic motion which utilizes a combination of a 4.0 kg mass and two identical springs mounted on a smooth horizontal table. each spring has a force constant of 75 n/m and an unstretched length of 0.25 m. the motion is started by releasing the mass from point a, where both springs have lengths of 0.5 m. one cycle is carried out as the mass moves from a to b to c to b to a. from the information given, find the speed of the mass as it passes through the central point b. 139 mls

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LectureNotes said a physics teacher designs a rather unusual demonstration of simple harmonic motion which utilizes a combination of a 4.0 kg mass and two identical springs mounted on a smooth horizontal table. each spring has a force constant of 75 n/m and an unstretched length of 0.25 m. the motion is started by releasing the mass from point a, where both springs have lengths of 0.5 m. one cycle is carried out as the mass moves from a to b to c to b to a. from the information given, find the speed of the mass as it passes through the central point b. 139 mls

Answer:

To find the speed of the mass as it passes through the central point ( B ), we need to use the principles of energy conservation in simple harmonic motion. Here’s a step-by-step approach:

1. Calculate the Potential Energy Stored in the Springs at Point A

When the mass is at point A, both springs are stretched from their natural length of 0.25 m to 0.5 m. The potential energy stored in a spring is given by:

U = \frac{1}{2} k x^2

where ( k ) is the spring constant and ( x ) is the displacement from the natural length.

For each spring:
U_{\text{spring}} = \frac{1}{2} \times 75 \, \text{N/m} \times (0.5 \, \text{m} - 0.25 \, \text{m})^2

U_{\text{spring}} = \frac{1}{2} \times 75 \, \text{N/m} \times (0.25 \, \text{m})^2

U_{\text{spring}} = \frac{1}{2} \times 75 \times 0.0625

U_{\text{spring}} = \frac{1}{2} \times 75 \times 0.0625

U_{\text{spring}} = 2.34375 \, \text{J}

Since there are two identical springs, the total potential energy at point A is:
U_{\text{total}} = 2 \times 2.34375 \, \text{J}

U_{\text{total}} = 4.6875 \, \text{J}

2. Conservation of Energy

At point B (the central point), all the potential energy will be converted into kinetic energy since it’s the equilibrium position where the springs are neither stretched nor compressed.

The kinetic energy ( K ) at point B is given by:

K = \frac{1}{2} m v^2

where ( m ) is the mass and ( v ) is the velocity.

Since the total mechanical energy is conserved, we have:
U_{\text{total}} = K

4.6875 \, \text{J} = \frac{1}{2} \times 4.0 \, \text{kg} \times v^2

3. Solve for the Velocity ( v )

4.6875 \, \text{J} = 2.0 \, \text{kg} \times v^2

v^2 = \frac{4.6875 \, \text{J}}{2.0 \, \text{kg}}

v^2 = 2.34375 \, \text{m}^2/\text{s}^2

v = \sqrt{2.34375 \, \text{m}^2/\text{s}^2}

v \approx 1.53 \, \text{m/s}

Therefore, the speed of the mass as it passes through the central point ( B ) is approximately ( 1.53 , \text{m/s} ). The value of 139 m/s seems to be incorrect based on the given data and calculations.