Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

Physics: Please help! I've done this 11 times!!!!!?

Attachment image

2 Answers

Relevance
  • hfshaw
    Lv 7
    2 months ago

    Assuming a "Hookian" spring for which the force is proportional to the displacement from its equilibrium position:

    F = - k*x

    where k is the spring constant (given as 197 N/m here), and x is the displacement from the equilibrium position.

    From Newton's second law:

    F = m*a = -k*x

    where m is the mass at the end of the (massless) spring, and a is its acceleration.

    The acceleration is the second derivative of the displacement, and the first derivative of the velocity:

    a = d²x/dt² = dv/dt = -(k/m)*x

    d²x/dt² + (k/m)*x = 0

    This is the equation of an undamped, unforced, harmonic oscillator. The solution is of the form:

    x(t) = a*cos(ω*t) + b*sin(ω*t)

    where ω = sqrt(k/m). In this case ω = sqrt((197 N/m)/(2.4 kg)) = 9.06/sec

    Differentiating this solution with respect to time gives us the equation for the velocity as a function of time, and differentiating again gives us the acceleration:

    v(t) = dx(t)/dx = -(a*ω)*sin(ω*t) + (b*ω)*cos(ω*t)

    a(t) = dv(t)/dt = -(a*ω²)*cos(ω*t) - (b*ω²)*sin(ω*t)

    We can now use the initial conditions for the position and velocity to determine the constants of integration, a and b:

    x(0) = 19 cm = a*cos(0) + b*sin(0) = a

    a = 19 cm

    v(0) = -29 cm/sec = -(a*ω)*sin(0) + (b*ω)*cos(0) = b*ω

    b = (-29 cm/sec)/(9.06/sec) = -3.2 cm

    So the acceleration as a function of time is given by:

    a(t) = -(19 cm)*(9.06/sec)²*cos(9.06*t/sec) - (3.2 cm)*(9.06/sec)²*sin(9.06*t/sec)

    At t = 2 sec, the acceleration is:

    a(2 sec) = -(19 cm)*(9.06/sec)²*cos(9.06*2) - (3.2 cm)*(9.06/sec)²*sin(9.06*2)

    a(2 sec) = -987.5 cm/s² = Assuming a "Hookian" spring for which the force is proportional to the displacement from its equilibrium position:

    F = - k*x

    where k is the spring constant (given as 197 N/m here), and x is the displacement from the equilibrium position.

    From Newton's second law:

    F = m*a = -k*x

    where m is the mass at the end of the (massless) spring, and a is its acceleration.

    The acceleration is the second derivative of the displacement, and the first derivative of the velocity:

    a = d²x/dt² = dv/dt = -(k/m)*x

    d²x/dt² + (k/m)*x = 0

    This is the equation of an undamped, unforced, harmonic oscillator. The solution is of the form:

    x(t) = a*cos(ω*t) + b*sin(ω*t)

    where ω = sqrt(k/m). In this case ω = sqrt((197 N/m)/(2.4 kg)) = 9.06/sec

    Differentiating this solution with respect to time gives us the equation for the velocity as a function of time, and differentiating again gives us the acceleration:

    v(t) = dx(t)/dx = -(a*ω)*sin(ω*t) + (b*ω)*cos(ω*t)

    a(t) = dv(t)/dt = -(a*ω²)*cos(ω*t) - (b*ω²)*sin(ω*t)

    We can now use the initial conditions for the position and velocity to determine the constants of integration, a and b:

    x(0) = 19 cm = a*cos(0) + b*sin(0) = a

    a = 19 cm

    v(0) = -29 cm/sec = -(a*ω)*sin(0) + (b*ω)*cos(0) = b*ω

    b = (-29 cm/sec)/(9.06/sec) = -3.2 cm

    So the acceleration as a function of time is given by:

    a(t) = -(19 cm)*(9.06/sec)²*cos(9.06*t/sec) - (3.2 cm)*(9.06/sec)²*sin(9.06*t/sec)

    At t = 2 sec, the acceleration is:

    a(2 sec) = -(19 cm)*(9.06/sec)²*cos(9.06*2) - (3.2 cm)*(9.06/sec)²*sin(9.06*2)

    a(2 sec) = -987.5 cm/s² = Assuming a "Hookian" spring for which the force is proportional to the displacement from its equilibrium position:

    F = - k*x

    where k is the spring constant (given as 197 N/m here), and x is the displacement from the equilibrium position.

    From Newton's second law:

    F = m*a = -k*x

    where m is the mass at the end of the (massless) spring, and a is its acceleration.

    The acceleration is the second derivative of the displacement, and the first derivative of the velocity:

    a = d²x/dt² = dv/dt = -(k/m)*x

    d²x/dt² + (k/m)*x = 0

    This is the equation of an undamped, unforced, harmonic oscillator. The solution is of the form:

    x(t) = a*cos(ω*t) + b*sin(ω*t)

    where ω = sqrt(k/m). In this case ω = sqrt((197 N/m)/(2.4 kg)) = 9.06/sec

    Differentiating this solution with respect to time gives us the equation for the velocity as a function of time, and differentiating again gives us the acceleration:

    v(t) = dx(t)/dx = -(a*ω)*sin(ω*t) + (b*ω)*cos(ω*t)

    a(t) = dv(t)/dt = -(a*ω²)*cos(ω*t) - (b*ω²)*sin(ω*t)

    We can now use the initial conditions for the position and velocity to determine the constants of integration, a and b:

    x(0) = 19 cm = a*cos(0) + b*sin(0) = a

    a = 19 cm

    v(0) = -29 cm/sec = -(a*ω)*sin(0) + (b*ω)*cos(0) = b*ω

    b = (-29 cm/sec)/(9.06/sec) = -3.2 cm

    So the acceleration as a function of time is given by:

    a(t) = -(19 cm)*(9.06/sec)²*cos(9.06*t/sec) - (3.2 cm)*(9.06/sec)²*sin(9.06*t/sec)

    At t = 2 sec, the acceleration is:

    a(2 sec) = -(19 cm)*(9.06/sec)²*cos(9.06*2) - (3.2 cm)*(9.06/sec)²*sin(9.06*2)

    a(2 sec) = -987.5 cm/s² = -9.9 m/s² (to the appropriate number of significant figures).

    • Login to reply the answers
  • Mangal
    Lv 4
    2 months ago

    Please see image....

    Attachment image
    • Login to reply the answers
Still have questions? Get your answers by asking now.