Apiece 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?

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

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

so here we can say

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

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.