A 190 g air-track glider is attached to a spring. The glider is pushed in 9.20 cm and released. A student with a stopwatch finds that 14.0 oscillations take 11.0 s. What is the spring constant?

Respuesta :

Answer:

[tex]k = 12.136\,\frac{N}{m}[/tex]

Explanation:

The angular frequency of the system is:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

The frequency is:

[tex]f = \frac{14\,osc}{11\,s}[/tex]

[tex]f = 1.272\,hz[/tex]

The angular frequency is:

[tex]\omega = 2\pi\cdot (1.272\,hz)[/tex]

[tex]\omega = 7.992\,\frac{rad}{s}[/tex]

The spring constant is:

[tex]k = \left(7.992\,\frac{rad}{s} \right)^{2}\cdot (0.190\,kg)[/tex]

[tex]k = 12.136\,\frac{N}{m}[/tex]