Physics questions from lab Hooke's Law and Spring Constant **Questions**1. Spring constant from the static method has units N/m, whereas in the dynamics method we get kg/s². Prove 1 N/m = 1 kg/s².2. How does the spring oscillate to be slower or faster on the Moon or on Jupiter, where the gravitational acceleration is much smaller and much larger respectively?3. How does the period of oscillation in dynamic methods change with increasing the mass?

We’ll answer the first question since the exact one wasn’t specified. Please submit a new question specifying the one you’d like answered. Static method for determining the spring constant In this method, a spring is taken and the position of the unstretched spring is noted. Let this be taken as x. We will get weights of known value for this experiment. Let the value of the weights be m1, m2, ..., mn. When attached to the spring these weights will stretch the spring and the position of the spring is noted for each weight be x1, x2, x3,..., xn. Now elongation for each weight will be calculated as ∆xn=xn-x The set to values for ∆x is plotted with respect to the masses and a straight line is obtained and its slope is calculated. Now for the equation of motion, we have the weight of the mass will be exactly equal to the force exerted by the spring. Thus mg=kx k is the spring constant and is given by k=mgx=gxm=gSlope of the graph Dynamic method for determining the spring constant In this method, the spring is set into oscillation with an applied mass. The differential equation of motion is Md2xdt2+kx=0 Here x is the elongation and M is the total mass of the spring and the weight. Therefore the differential equation can be written as d2xdt2+kMx=0 Thus we have ω2=kM⇒ω=kM⇒2πT=kM⇒k=4π2T2M The time period of the oscillations is noted for every value of M and a plot is drawn which shows T2 vs M and the slope of the graph is noted. k=4π2T2M=4π2T2M=4π2Slope of the graph In the static method, we get the value of spring constant in a unit of N/m and in the dynamic method we get the unit of spring constant as kg/s2. 1 N/m is the constant of spring if a force of 1 N elongates the spring by 1 m. From the Newton's second law motion we get F=ma F, m and a are the force, mass and accelerations respectively. Thus the value of 1 N is 1 N=1 kg. 1 m/s2=1 kgm/s2 Thus we have 1 N/m=1 kgm/ms2⇒1 N/m=1kg/s2Answer: We have successfully proved 1 N/m=1 kg/m2.

Answered: 1. Spring constant from the static…

1. Spring constant from the static method has units N/m, whereas the dynamics method we get kg/s². Prove 1N/m= 1kg/s².