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A spring is compressed by 0.0620m and is used to launch an object horizontally with a speed of 1.50(m)/(s). If the object were attached to the spring, at what angular frequency (in ra)?
A spring is compressed by 0.0620m and is used to launch an object horizontally with a speed of 1.50(m)/(s). If the object were attached to the spring, at what angular frequency (in ra)?
The angular frequency of the spring, if the object were attached to it, is approximately 12.12 rad/s.When the object is launched horizontally from the compressed spring, it experiences projectile motion. However, if the object were attached to the spring, it would undergo simple harmonic motion (SHM) due to the restoring force exerted by the spring. The angular frequency (ω) of an object undergoing SHM with an amplitude A and period T is given by the formula ω = 2π / T = 2πf, where f is the frequency of oscillation.To find the angular frequency in this case, we need to relate the given information to the equations of motion for SHM. The amplitude A of the motion is the displacement of the spring when compressed, given as 0.0620 m. The velocity of the object when launched horizontally from the spring is 1.50 m/s.We know that the maximum velocity of an object undergoing SHM is ωA. Therefore, we can use this information along with the given velocity to find the angular frequency. Rearranging the formula, we get ω = v / A. Substituting the values, we get ω ≈ 1.50 / 0.0620 ≈ 24.19 rad/s....