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Olin [163]
3 years ago
7

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 g

um 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?
Physics
1 answer:
cestrela7 [59]3 years ago
7 0

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

y = R sin\theta

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

\theta = \omega t

so here we can say

y = R sin(\omega t)

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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