Question

A 2.20 kg frictionless block is attached to an ideal spring with force constant 314 n/m . Initially the block has velocity -3.70 m/s and displacement 0.270 m.

Find the amplitude of the motion in m

Find the maximum acceleration of the block in m/s^2

Find the maximum force the spring exerts on the block in n

Answer 1

Concepts and reason

The concept of simple harmonic motion is required to solve the problem.

First, determine the angular velocity of the motion by using the relation between angular velocity, spring constant, and mass of the block. Then determine the amplitude of the motion by using the expression for the velocity of the block. The maximum acceleration of the block is determined by using the relation between angular velocity, amplitude, and acceleration of the block. The maximum force is determined by using the relation between spring constant, amplitude, and force.

Fundamentals

The velocity of the block undergoing simple harmonic motion is given as,

$v = \omega \sqrt {{A^2} - {x^2}}$

Here, A is the amplitude of the motion, $\omega$ is the angular velocity, and x is the position of the block.

The angular velocity is given by the following relation:

$\omega = \sqrt {\frac{k}{m}}$

Here, k is the spring constant and m is the mass of the block.

The maximum acceleration of the block is given as,

$a = {\omega ^2}A$

Here, A is the amplitude of the motion.

The maximum force exerted on the block is given as,

$F = kA$

Here, k is the spring constant and A is the amplitude of the motion.

(a)

Determine the amplitude of motion.

The velocity of the block is given as,

$v = \omega \sqrt {{A^2} - {x^2}}$

The angular velocity is given as,

$\omega = \sqrt {\frac{k}{m}}$

Here, k is the spring constant and m is the mass of the block.

Substitute $\sqrt {\frac{k}{m}}$ for $\omega$ in equation $v = \omega \sqrt {{A^2} - {x^2}}$ and rearrange the equation for A.

$\begin{array}{c}\\v = \sqrt {\frac{k}{m}\left( {{A^2} - {x^2}} \right)} \\\\{v^2} = \frac{k}{m}\left( {{A^2} - {x^2}} \right)\\\\A = \sqrt {\frac{{m{v^2}}}{k} + {x^2}} \\\end{array}$

Substitute 2.20 kg for m, -3.70 m/s for v, 314 N/m for k, and 0.270 m for x in the equation $A = \sqrt {\frac{{m{v^2}}}{k} + {x^2}}$ and determine the amplitude of the motion.

$\begin{array}{c}\\A = \sqrt {\frac{{\left( {2.20{\rm{ kg}}} \right){{\left( { - 3.70{\rm{ m/}}{{\rm{s}}^2}} \right)}^2}}}{{314{\rm{ N/m}}}} + {{\left( {0.270{\rm{ m}}} \right)}^2}} \\\\ = 0.411{\rm{ m}}\\\end{array}$

(b)

The maximum acceleration of the block is given as,

$a = {\omega ^2}A$

Here, $\omega$ is the angular velocity and A is the amplitude of motion.

Substitute $\sqrt {\frac{k}{m}}$ for $\omega$ in the above equation.

$a = \frac{k}{m}A$

Substitute 2.20 kg for m, 314 N/m for k, and 0.411 m for A in the equation $a = \frac{k}{m}A$ and determine the maximum acceleration of the block.

$\begin{array}{c}\\a = \frac{{314{\rm{ N/m}}}}{{2.20{\rm{ kg}}}}\left( {0.411{\rm{ m}}} \right)\\\\ = 58.7{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

(c)

Calculate the maximum force the spring exerts on the block.

The maximum force exerted by the spring on the block is,

$F = kA$

Here, k is the spring constant and A is the amplitude of motion.

Substitute 314 N/m for k and 0.411 m for A in the equation $F = kA$.

$\begin{array}{c}\\F = \left( {314{\rm{ N/m}}} \right)\left( {0.411{\rm{ m}}} \right)\\\\ = 129{\rm{ N}}\\\end{array}$

Ans: Part a

The amplitude of motion is 0.411 m.

Part b

The maximum acceleration of the block is $58.7{\rm{ m/}}{{\rm{s}}^2}$.

Part c

The maximum force the spring exerts on the block is 129 N.