Q10Q You are to build the oscillation... [FREE SOLUTION]

Chapter 15: Q10Q (page 435)

You are to build the oscillation transfer device shown in Fig.15-27. It consists of two spring鈥揵lock systems hanging from a flexible rod. When the spring of system is stretched and then released, the resulting SHM of system at frequency oscillates the rod. The rod then exerts a driving force on system 2, at the same frequency $f_{1}$ . You can choose from four springs with spring constants k of 1600,1500,1400, and 1200 N/m, and four blocks with masses m of 800,500,400, and 200 kg. Mentally determine which spring should go with which block in each of the two systems to maximize the amplitude of oscillations in system 2.

Short Answer

Expert verified

Pair of spring block system to maximize amplitude of oscillations in system 2 is for the first system spring constant is 1500N/m and mass is 500kg while for the second system spring constant is 1200N/m and mass is 400kg.

Step by step solution

The given data

Figure for given system. The values of the spring constants and masses are:

$k_{1} = \frac{1600 N}{m}, k_{2} = \frac{1500 N}{m}, k_{3} = \frac{1400 N}{m} and k_{4} = \frac{1200 N}{m} m_{1} = 800 kg, m_{2} = 500 kg, m_{3} = 400 kg, m_{4} = 200 kg$

Understanding the concept of frequency in SHM

We can use the formula of frequency which is related to mass and spring constant. To maximize the amplitude in the second system both systems should have equal frequencies. Using this relation, we can check which mass and spring constant pair that satisfies the relation.

Formula:

The frequency of the oscillation in SHM, $f = \frac{1}{2 蟺} \sqrt{\frac{k}{m}}$ (i)

Calculation of the spring constants and mass for the spring-block system

We know for maximum amplitude of oscillation in system 2; it should have frequency equal to frequency of system 1.

$f_{1} = f_{2}$

From equation (i), we can get the frequency equation as follows:

$\frac{1}{2 蟺} \sqrt{\frac{k_{1}}{m_{1}}} = \frac{1}{2 蟺} \sqrt{\frac{k_{2}}{m_{2}}} \sqrt{\frac{k_{1}}{m_{1}}} = \sqrt{\frac{k_{2}}{m_{2}}} \frac{k_{1}}{m_{1}} = \frac{k_{2}}{m_{2}}$

We can check the above equation is satisfied for the given masses and spring constants, that is

These values satisfy the above equation.

Hence, the pair of spring constant and mass with the values for first system spring constant is 1500N/m and mass is, while for second system spring constant is 1200N/m and mass is 400kg.