A 0.40 kg mass vibrates at the end of a horizontal spring along a frictionless surface reaching a maximum speed of 0.50m/s If the maximum displacement of the mass is 0.11 m, what is the spring constant? (8.3N/m)

Here's how to solve this problem:<br /><br />**1. Understand the relationship between maximum speed, maximum displacement, and spring constant.**<br /><br />In simple harmonic motion (SHM) with a spring, the maximum speed (v_max) and maximum displacement (A) are related to the spring constant (k) and mass (m) by the following equation:<br /><br />v_max = √(k/m) * A<br /><br />**2. Rearrange the equation to solve for k:**<br /><br />k = (v_max² * m) / A²<br /><br />**3. Plug in the given values:**<br /><br />* v_max = 0.50 m/s<br />* m = 0.40 kg<br />* A = 0.11 m<br /><br />k = (0.50² * 0.40) / 0.11²<br /><br />**4. Calculate the spring constant:**<br /><br />k = (0.25 * 0.40) / 0.0121<br />k = 0.1 / 0.0121<br />k ≈ 8.26 N/m<br /><br />**5. Round to the appropriate number of significant figures:**<br /><br />Since the given values have two significant figures, the answer should also have two significant figures.<br /><br />k ≈ 8.3 N/m<br /><br />Therefore, the spring constant is approximately 8.3 N/m.<br />