In a horizontal spring-mass configuration, the position of the 0.52 kg mass is described by [x = (1.09 m) sin(2.70t)]
a) What is in meters the amplitude of these oscillations?
b) What is in m/s the maximum speed of the mass under these conditions?
c) What is in Joules the spring potential energy stored in this system when the spring is stretched at a maximum?
d) What is in Joules the kinetic energy of the mass when the spring is not stretched?
e) What is in Joules the kinetic energy of the mass when the spring is stretched by 0.22 m?
To answer these questions, we need to understand the properties of a spring-mass system and its equations of motion. Let's break down each question and explain how to find the answers step by step:
a) The amplitude of oscillations in a spring-mass system represents the maximum displacement from the equilibrium position. In this case, the equation of motion is given as x = (1.09 m) sin(2.70t). The amplitude can be identified as the coefficient of the sine function, which is 1.09 m. Therefore, the amplitude of these oscillations is 1.09 meters.
b) The maximum speed of the mass can be found by taking the derivative of the displacement equation with respect to time (t). The derivative of sin(2.70t) is 2.70 cos(2.70t). So, the derivative of the displacement equation is dx/dt = (1.09 m)(2.70 cos(2.70t)). To find the maximum speed, we need to evaluate the derivative at the extreme points of the motion, which occur when cos(2.70t) is at its maximum value of 1. Therefore, the maximum speed is (1.09 m)(2.70 × 1) = 2.943 m/s.
c) The spring potential energy is given by the equation U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position. In this case, the equation for displacement is x = (1.09 m) sin(2.70t), and when the spring is stretched at its maximum, x = 1.09 m. The spring constant (k) is not given, so it needs to be provided to calculate the spring potential energy (U).
d) When the spring is not stretched, the displacement (x) is zero. Therefore, the potential energy (U) is also zero because U = (1/2)kx^2. However, to calculate the kinetic energy (KE) of the mass, we need its velocity. Given that the spring is not stretched, the mass is at its equilibrium position, where the velocity is zero. Therefore, the kinetic energy of the mass is also zero.
e) The kinetic energy (KE) of the mass can be calculated using the equation KE = (1/2)mv^2, where KE is kinetic energy, m is the mass, and v is the velocity. The velocity can be found by differentiating the displacement equation with respect to time (t). So, dx/dt = (1.09 m)(2.70 cos(2.70t)). To calculate the kinetic energy when the spring is stretched by 0.22 m, we need to plug this value into the displacement equation, x = (1.09 m) sin(2.70t), and find the time (t) when x = 0.22 m. Once we find the time, we can calculate the velocity (v) using the derivative we previously computed. Finally, we can use the velocity and mass values to calculate the kinetic energy (KE).
Please provide the value of the spring constant (k) to proceed with the calculations for part c and e.