Question

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 1

T = 2.82 seconds.

The frequency
`\mathbf{f = 0.36 \ Hz}`

Amplitude A = 25.5 cm

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

Step-by-step explanation:

Given that:

the time taken for 11 oscillations is 31 seconds ;

SO, the time taken for one oscillation is :

`T = (31)/(11)`

T = 2.82 seconds.

The formula for calculating frequency can be expressed as :

`f = (1)/(T)`

`f = (1)/(2.82)`

`\mathbf{f = 0.36 \ Hz}`

The amplitude is determined by using the formula:

`A = (d)/(2)`

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

`A = (51)/(2)`

A = 25.5 cm

The maximum speed of the glider is:

`v = A \omega`

where ;

`\omega = (2 \pi)/(T)`

`\omega = (2 \pi)/(2.82)`

`\omega = 2.23 \ rad/s`

`v = A \omega`

`v = 25.5 *2.23`

`\mathbf{v = 56.87 \ rad/s}`