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ikadub [295]
3 years ago
6

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 compl

etes 11.0 oscillations in 31.0 s
Physics
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

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}

Explanation:

Given that:

the time taken for 11 oscillations is 31 seconds ;

SO, the time taken for one oscillation is :

T = \frac{31}{11}

T = 2.82 seconds.

The formula for calculating frequency can be expressed as :

f = \frac{1}{T}

f = \frac{1}{2.82}

\mathbf{f = 0.36 \ Hz}

The amplitude is determined by using the formula:

A = \frac{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 = \frac{51}{2}

A = 25.5 cm

The maximum speed of the glider is:

v = A \omega

where ;

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

\omega = \frac{2 \pi}{2.82}

\omega = 2.23 \ rad/s

v = A \omega

v = 25.5 *2.23

\mathbf{v = 56.87 \ rad/s}

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