Question

In: Physics

An air-track glider is attached to a spring. The glider is pulled to the right and...

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?

Solutions

Expert Solution

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

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

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

T=2πωT = \frac{{2\pi }}{\omega }

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

v=Aωv = A\omega

(A)

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

T=2πωT = \frac{{2\pi }}{\omega }

Rearrange the expression.

ω=2πT\omega = \frac{{2\pi }}{T}

Substitute 1.50 s for T.

ω=2π1.50s=4.189rad/s\begin{array}{c}\\\omega = \frac{{2\pi }}{{1.50{\rm{ s}}}}\\\\ = 4.189{\rm{ rad/s}}\\\end{array}

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

v=Aωv = A\omega

Rearrange the expression for the amplitude of the oscillation.

A=vωA = \frac{v}{\omega }

Substitute 4.189s14.189{\rm{ }}{{\rm{s}}^{ - 1}} for ω\omega and 50.0 cm/s for v.

A=50cm/s4.189s1=11.94cm\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(t)=Acos(ωt)x\left( t \right) = A\cos \left( {\omega t} \right)

Substitute 11.94 cm for A, 29 s for t, and 4.189rad/s4.189{\rm{ rad/s}} for ω\omega .

x(t)=(11.94cm)cos((4.189s1)(29s))=6.0cm\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 6.0cm- 6.0{\rm{ cm}} .


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