contestada

A piece of gum is stuck to the outer edge of a horizontal turntable, which is revolving at a constant speed. The shadow of the gum is projected onto a wall behind the turntable. How does the frequency of rotation of the turntable compare to the frequency of oscillation of the gum's shadow?

Respuesta :

Answer:

angular frequency of the table must be same as the frequency of the projection of the gum on the wall

Explanation:

Since we know that the projection on the wall is the vertical component of the position of the gum on the rotating table

So here we will say

[tex]y = R sin\theta[/tex]

so the angle made by the radius vector depends on the angular frequency of the disc by which it is rotating

So we can say

[tex]\theta = \omega t [/tex]

so here we can say

[tex]y = R sin(\omega t)[/tex]

so here we can say that

angular frequency of the table must be same as the frequency of the projection of the gum on the wall

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