An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 61.0 cm mark on the track. The glider completes 11.0 oscillations in 31.0 s

Answer:

T = 2.82 seconds.

The frequency [tex]\mathbf{f = 0.36 \ Hz}[/tex]

Amplitude A = 25.5 cm

The maximum speed of the glider is [tex]\mathbf{v = 56.87 \ rad/s}[/tex]

Explanation:

Given that:

the time taken for 11 oscillations is 31 seconds ;

SO, the time taken for one oscillation is :

[tex]T = \frac{31}{11}[/tex]

T = 2.82 seconds.

The formula for calculating frequency can be expressed as :

[tex]f = \frac{1}{T}[/tex]

[tex]f = \frac{1}{2.82}[/tex]

[tex]\mathbf{f = 0.36 \ Hz}[/tex]

The amplitude is determined by using the formula:

[tex]A = \frac{d}{2}[/tex]

The limits that the spring makes the oscillations are from 10 cm to 61 cm.

The distance of the glider is, d = (61 - 10 )cm = 51 cm

Replacing 51 for d in the above equation

[tex]A = \frac{51}{2}[/tex]

A = 25.5 cm

The maximum speed of the glider is:

[tex]v = A \omega[/tex]

where ;

[tex]\omega = \frac{2 \pi}{T}[/tex]

[tex]\omega = \frac{2 \pi}{2.82}[/tex]

[tex]\omega = 2.23 \ rad/s[/tex]

[tex]v = A \omega[/tex]

[tex]v = 25.5 *2.23[/tex]

[tex]\mathbf{v = 56.87 \ rad/s}[/tex]

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