Question

    A mass m at the end of a spring vibrates with a frequency of 0.90 Hz. When an additional 520-g mass is added to m, the frequency is 0.50 Hz.  

   a) What is the value of m?                                                                                                

  b) Calculate the total mass needed for the system to vibrate with a freq. of 0.45 Hz. 

Answer 1

$f = \dfrac{1}{2π}\sqrt{\dfrac{k}{m}}$

$k =4π²mf²$

$\therefore 4π²m_1f_1² = 4π²m_2f_2²$

$m_1f_1² = m_2f_2²$

a.

$m_1× 0.90² = (m_1+520) × 0.50²$

$0.90² m_1= 0.50²m_1 +130$

$0.56m_1 = 130$

$m_1 =232.14g$

b.

$k =4π²mf²$

$k = 4π²×232.14 ×0.90 = 8248.07Nm^{-1}$

$m= \dfrac{k}{4π²f²} =\dfrac{8248.07}{4π²×0.45²} =$

$1031.73g \approx1.031kg$