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stepladder [879]
2 years ago
12

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. the mass is pull

ed away from the equilibrium position (x = 0) a distance of 17.5 cm and released. it then oscillates in simple harmonic motion with a frequency of 8.38 hz. at what position, measured from the equilibrium position, is the mass 2.50 seconds after it is released?
Physics
2 answers:
White raven [17]2 years ago
6 0
For a simple harmonic motion, the position of the mass at any time t is given by
x(t)=A cos (\omega t)
where 
A is the amplitude of the motion (in this problem, A=17.5 cm)
\omega is the angular frequency of the oscillator
t is the time

The angular frequency of the motion in the problem is given by
\omega = 2 \pi f= 2 \pi (8.38 Hz) = 52.7 rad/s

And so, we can find the position x of the mass (with respect to the equilibrium position) at time t=2.50 s:
x(2.50 s)=(17.5 cm) \cos ( (52.7 rad/s)(2.50 s))=17.2 cm
Dmitriy789 [7]2 years ago
6 0

16.64cm is the correct answer.

Don't round ω when entering cos(ω*T) in your calculator


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v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s

D) 9.3 m/s

The y-component of the shot's velocity is given by:

v_y = v_0 sin \theta

where

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E) 7.6 m/s

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Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

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2 years ago
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