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

General
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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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Questionmatic 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 m/s

Answer:

To solve this problem, we will use the principles of energy conservation. When the mass is released from point A, it has potential energy stored in the springs. As it passes through point B, this potential energy gets converted into kinetic energy.

  1. Calculate the potential energy stored in the springs at point A:

    At point A, both springs are stretched from their natural length of 0.25 m to 0.5 m. The extension (\Delta x) in each spring is:

    \Delta x = 0.5 \ \text{m} - 0.25 \ \text{m} = 0.25 \ \text{m}

    The potential energy (U_s) stored in one spring is given by:

    U_s = \frac{1}{2} k (\Delta x)^2

    where k = 75 \ \text{N/m} (force constant of the springs).

    Therefore, for one spring:

    U_s = \frac{1}{2} \times 75 \ \text{N/m} \times (0.25 \ \text{m})^2
    U_s = \frac{1}{2} \times 75 \times 0.0625
    U_s = \frac{1}{2} \times 75 \times 0.0625 = \frac{75}{2} \times 0.0625 = 37.5 \times 0.0625 = 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} = 4.6875 \ \text{J}

  2. Energy conservation:

    At point B, the springs are neither compressed nor stretched, which means all the potential energy from point A is converted into kinetic energy (K) at point B. Therefore:

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

    The kinetic energy of the mass is given by:

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

    where m = 4.0 \ \text{kg} (mass of the object) and v is the velocity we need to find. So,

    4.6875 \ \text{J} = \frac{1}{2} \times 4.0 \ \text{kg} \times v^2
    4.6875 = 2.0 \times v^2

    Dividing both sides by 2.0:

    v^2 = \frac{4.6875}{2.0}
    v^2 = 2.34375

    Taking the square root of both sides:

    v = \sqrt{2.34375}
    v \approx 1.53 \ \text{m/s}

Final Answer:
The speed of the mass as it passes through the central point B is approximately \boxed{1.53 \ \text{m/s}}.