A mass m at the end of a spring oscillates with a frequency of 0.83 Hz. When an additional 780-g mass is added to m, the frequency is 0.60 Hz. What is the value of m?
T1=1ν1=2πmk→10.83=2πmkT_1=\frac{1}{\nu_1}=2\pi\sqrt{\frac{m}{k}}\to\frac{1}{0.83}=2\pi\sqrt{\frac{m}{k}}T1=ν11=2πkm→0.831=2πkm
T2=1ν2=2πm+0.78k→10.6=2πm+0.78kT_2=\frac{1}{\nu_2}=2\pi\sqrt{\frac{m+0.78}{k}}\to\frac{1}{0.6}=2\pi\sqrt{\frac{m+0.78}{k}}T2=ν21=2πkm+0.78→0.61=2πkm+0.78
0.60.83=mm+0.78→0.52257=mm+0.78→m=0.854 (kg)=854 g\frac{0.6}{0.83}=\sqrt{\frac{m}{m+0.78}}\to0.52257=\frac{m}{m+0.78}\to m=0.854\ (kg)=854\ g0.830.6=m+0.78m→0.52257=m+0.78m→m=0.854 (kg)=854 g . Answer
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