Question

A block with mass m is undergoing shm on a horizontal, frictionless surface while attached to a light, horizontal spring that has force constant k . you use motion sensor equipment to measure the maximum speed of the block during its oscillations. you repeat the measurement for the same spring and blocks of different masses while keeping the amplitude a at a constant value of 14.0 cm . you plot your data as v2max versus 1/m and find that the data lie close to a straight line that has slope 7.62 n⋅m .

Answer 1

The question is about the relation between maximum velocity squared and the inverse of mass in simple harmonic motion, which can be derived from the equation vmax = A sqrt(k/m), revealing that the slope of the v2max versus 1/m graph equals kA2.

Step-by-step explanation:

The student's question pertains to the simple harmonic motion (SHM) of a block attached to a spring on a frictionless surface and how this motion relates to different masses. They have plotted maximum velocity squared (v2max) against the inverse of mass (1/m) and found a linear relationship. This experiment involves concepts of energy conservation in SHM, kinetic energy, and potential energy of springs. The slope of the graph obtained from plotting the data points represents the product of the amplitude squared and the spring constant (a2k).

To understand the problem, we can start with the equation for maximum velocity in SHM, which is vmax = Aω, where A is the amplitude and ω is the angular frequency. Since ω = sqrt(k/m) for a mass-spring system, it follows that vmax = Aω = A sqrt(k/m). This implies that v2max is proportional to kA2/m, hence when graphed against 1/m the slope is kA2, from which the spring constant k can be determined if A is known.