Calculate the angular frequency of the system shown in the figure. Friction is absent and threads are massless? Hint: The system performs SHM. Try not to use the energy conservation method.

#m_A = m_B = m#

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1 Answer
May 18, 2018

Let #x_0# be the extension in the spring having spring constant #k# while in equilibrium condition. If #T# is tension in one string, then equilibrium of masses #A and B# respectively give the following force equations

#kx_0+m_Agsintheta=T# ..........(1)
#2T=m_Bg# ..........(2)

Rewriting (1) with the help of (2)

#kx_0+m_Agsintheta=(m_Bg)/2# .....(3)

Let the spring be stretched by a displacement #x# from the mean position where #v# be velocity of mass #m_A#. As there is double string attached with mass #m_B#
1. Its velocity would be #=v/2#, and
2. it is lowered by a displacement #=x/2#
Total Energy equation of the system becomes

#TE="PE of spring"+"KE of mass "m_A+"KE of mass "m_B+"PE of "m_A+"PE of mass "m_B#
#=>TE=1/2k(x_0+x)^2+1/2m_Av^2+1/2m_B(v/2)^2+m_Agxsintheta-m_Bgx/2#
#=>TE=1/2k(x_0+x)^2+1/2m_Av^2+1/8m_Bv^2+(m_Agsintheta)x-(1/2m_Bg)x#

Differentiating with respect to time and noting that Law of conservation of energy holds so that #d/dt(TE)=0#, above equation becomes

#0=d/dt(1/2k(x_0+x)^2+1/2m_Av^2+1/8m_Bv^2+(m_Agsintheta)x-(1/2m_Bg)x)#
#=>0=k(x_0+x)dotx+m_Avdotv+1/4m_Bvdotv+(m_Agsintheta)dotx-(1/2m_Bg)dotx#

We know that #dotx=v and dotv=ddotx#. Using (3) and with these above equation simplifies to

#0=kx+m_Addotx+1/4m_Bddotx#

Given #m_A=m_B=m#. With this above becomes

#0=kx+mddotx+1/4mddotx#
#=>-kx=5/4mddotx#
#=>-kx=m^'ddotx# .......(4)
where #m^'=5/4m#

(4) is a second-order linear ordinary differential equation for which angular frequency is given as

#omega=sqrt(k/m^')#

Inserting value of #m^'# we get

#omega=sqrt(4/5k/m)#