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2 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

1 answer:

Novosadov [1.4K]2 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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