Question

An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at =0 s. It then oscillates with a period of 1.50s and a maximum speed of 50.0 cm/s .

What is the amplitude of the oscillation?

What is the glider's position a t = 29.0s?

Answer 1

Concepts and reason

The concepts required to solve this question are simple harmonic motion, displacement, and velocity.

Initially, write the expression of the time period. Then, rearrange the expression for the angular frequency. Using the expression of the maximum speed calculate the amplitude of the oscillation. Finally, using the expression of the general equation of the displacement for the oscillation calculate the position at time t.

Fundamentals

The general equation of the displacement for simple harmonic motion is,

$x\left( t \right) = A\cos \left( {\omega t} \right)$

Here, $x\left( t \right)$ is the displacement of the particle, A is the amplitude, t is the time, and $\omega$ is the angular frequency.

The expression of the time period $\left( T \right)$ is,

$T = \frac{{2\pi }}{\omega }$

The expression of the maximum speed $\left( v \right)$ is,

$v = A\omega$

(A)

The expression of the time period $\left( T \right)$ is,

$T = \frac{{2\pi }}{\omega }$

Rearrange the expression.

$\omega = \frac{{2\pi }}{T}$

Substitute 1.50 s for T.

$\begin{array}{c}\\\omega = \frac{{2\pi }}{{1.50{\rm{ s}}}}\\\\ = 4.189{\rm{ rad/s}}\\\end{array}$

The expression of the maximum speed $\left( v \right)$ is,

$v = A\omega$

Rearrange the expression for the amplitude of the oscillation.

$A = \frac{v}{\omega }$

Substitute $4.189{\rm{ }}{{\rm{s}}^{ - 1}}$ for $\omega$ and 50.0 cm/s for v.

$\begin{array}{c}\\A = \frac{{50{\rm{ cm/s}}}}{{4.189{\rm{ }}{{\rm{s}}^{ - 1}}}}\\\\ = 11.94{\rm{ cm}}\\\end{array}$

(B)

The general equation of the displacement for simple harmonic motion is,

$x\left( t \right) = A\cos \left( {\omega t} \right)$

Substitute 11.94 cm for A, 29 s for t, and $4.189{\rm{ rad/s}}$ for $\omega$ .

$\begin{array}{c}\\x\left( t \right) = \left( {11.94{\rm{ cm}}} \right)\cos \left( {\left( {4.189{\rm{ }}{{\rm{s}}^{ - 1}}} \right)\left( {29{\rm{ s}}} \right)} \right)\\\\ = - 6.0{\rm{ cm}}\\\end{array}$

Ans: Part A

The amplitude is 11.94 cm.

Part B

The glider’s position is $- 6.0{\rm{ cm}}$ .