What Causes Oscillation Errors in Air-Track Glider Experiments?

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Oscillation errors in air-track glider experiments can arise from incorrect calculations related to angular displacement. The glider, attached to a spring, oscillates with a period of 1.50 seconds and a maximum speed of 46.0 cm/s, with an amplitude of 11 cm. A user attempted to calculate the glider's position at t=26.0 seconds using the formula x = A cos(wt) but received an incorrect answer. The error was identified as a misconfiguration of the calculator, which was set to degrees instead of radians. Properly using radians in calculations is crucial for accurate results in oscillation problems.
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An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0. It then oscillates with a period of 1.50 s and a maximum speed of 46.0 cm/s.

I found the amplitude to be 11cm (this is 100% correct)

1) What is the glider's position at t= 26.0 s?

How do i do #1?

I used the formula x = A cos ( wt) and plugged everything in.

x = (11cm) cos (26sec x 2pie/1.5)
x = -3.56

but my answer is wrong :-(
 
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Your angle 26*2*pi/1.5 is in radians, but I think your calculator was set to work in degrees.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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