Question

4. [2/3 Points) DETAILS PREVIOUS ANSWERS SERCP11 13.4.P.020. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A student stretches a spring, attaches a 1.90 kg mass to it, and releases the mass from rest on a frictionless surface. The resulting oscillation has a period of 0.650 s and an amplitude of 30.0 cm. Determine the oscillation frequency, the spring constant, and the speed of the mass when it is halfway to the equilibrium position. HINT (a) the oscillation frequency (in Hz) 1.54 Remember there are two frequencies: the frequency, f, gives the number of cycles per second, whereas the angular frequency, w, gives the number of radians per second. These two physical concepts are nearly identical and are linked by the conversion factor 27 rad/cycle. Hz (b) the spring constant (in N/m) 177.36 N/m (c) the speed of the mass (in m/s) when it is halfway to the equilibrium position * m/s 2.89 Need Help? Read It Watch

Answer 1

total energy remains conserved. Maximum energy is 1/2(mw²A²). So, at mid point x= A/2. Kinetic energy 1/2(mv²). Potential energy = 1/2(mw²x²).

1.54 Hz. K= Spring Constant = 177.36 N/m f frequency = 1.54 M angular mara = 1.90 kg. 2 rif g. 676 rad/s. ma A= 30 cm - 0.30 m x= A² 12 mw x² + mv² = mwa I mwa A² + s mest = smu?n? morir gh wh = ².f. w?. A?, ah w* .(9-68) ** (0-3) m/s = 6.32 m/s. - 2.51 m/s. ARA nart our point.)