Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago Relevance
• 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).

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