Question

A 22.5 kg chair attached to a spring takes 1.30 s to complete one cycle of oscillation. With an astronaut sitting in the oscillating chair with feet off the floor, a full oscillation cycle takes 2.54 s. What is the astronaut's mass?

Answer 1

85.689 kg is the astronaut's mass.

Explanation:

Given data: m = 22.5 kg   and T = 1.3 sec

So, using the below formula,

                            $T = 2 \pi \sqrt{\frac{m}{k}}$

Now, after putting the values of m and T in the above equation, we will find out the value of k which is as,

                            $1.3 = 2 \times\left(\frac{22}{7}\right) \sqrt{\frac{22.5}{k}}$

                            $1.3 \times 7=44 \sqrt{\frac{22.5}{k}}$

                            $\frac{9.1}{44}=\sqrt{\frac{22.5}{k}}$

To remove the square root, take square on both sides, we get,

                           $(0.2068)^{2}=\frac{22.5}{k}$

                           $k=\frac{22.5}{0.0428}=525.7 \mathrm{N} / \mathrm{m}$

Now, we have the same string but this time we have different mass and different time. So, let the mass of the astronaut is $m=m_{a}$ and $T_{a}$ = 2.54 sec, k= 525.7 kg. Apply these values in the equation, we get,

                         $2.54 = 2 \times 3.14\left(\frac{m_{a}}{525.7}\right)^{\frac{1}{2}}$

                        $\frac{2.54}{6.28} = \left(\frac{m_{a}}{525.7}\right)^{\frac{1}{2}}$

                        $0.404 = \left(\frac{m_{a}}{525.7}\right)^{\frac{1}{2}}$

Taking squares on both sides, we get,

                        $m_{a} = 525.7 \times(0.404)^{2}=525.7 \times 0.163 = 85.689 \mathrm{kg}$