contestada

A 0.80 kg mass attached to an ideal spring oscillates horizontally with a period of 0.50 s and amplitude of 0.30 m. What is the spring constant of the spring?

Respuesta :

opudodennis

Answer:

126.33 N/m

Explanation:

Angular frquency is given by

[tex]f=\frac {2\pi}{t}[/tex]

Wheret is period of wave and f is angular frequency.

Given period of 0.5 s then

[tex]w=\frac {2\pi}{0.5}=4\pi rad/s\approx 12.57 rad/s[/tex]

Angular frequency is also given by

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

Making k the subject then [tex]k=mw^{2}[/tex]

Where k is spring constant, m is mass

Substituting m with 0.8 then

[tex]k=0.8\times (4\pi)^{2}[/tex]

K=126.330936333944 N/m

Rounded off as 126.33 N/m

Cricetus

The spring constant of the spring will be "31.58 N/m".

Given:

  • Mass = 0.8 kg
  • Time = 0.50 s

The angular frequency will be:

→ [tex]\omega = \frac{2 \pi}{T}[/tex]

      [tex]= \frac{2 \pi}{0.5}[/tex]

      [tex]= 4 \pi \ rad/s[/tex]

As we know,

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

hence,

The spring constant will be:

→ [tex]K = m \omega^2[/tex]

By substituting the values, we get

       [tex]= 0.8\times (4 \pi)^2[/tex]

       [tex]= 31.58 \ N/m[/tex]

Thus the approach above is correct.      

Learn more about amplitude here:

https://brainly.com/question/22209357

Otras preguntas