Logarithmic Decrement (​δ)

Logarithmic Decrement (​δ)

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fig_logarithmic-decrement fig-1: displacement V/s time curve for under damped system

Logarithmic decrement

Logarithmic decrement is defined as the natural logarithm of the ratio of successive amplitude on the same side of mean position.

The rate of decay in the amplitudes of under-damped system is measured by the parameter known as logarithmic decrement.

Rate of decay in amplitudes depends on the amount of damping present in the system. So if the damping is more, then the rate of decay will also be more.

Let A and B are the two points on the successive cycles which shows maximum deflection as shown in figure.

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The periodic time:

tp=t2t1=2πωd=2π(1ξ2) ωn\begin{aligned} t_p &= t_2 − t_1 \\ &= {2 \pi \over \omega_d} \\ &= {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \end{aligned}

The amplitude at time t1t_1 and t2t_2 are:

x1=Xeξωnt1[sin(ωdt1+)]x_1 = Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)]

And,

x2=Xeξωnt2[sin(ωdt2+)]x2=Xeξωn(t1+tp)[sin{ωd(t1+tp)+}]x2=Xeξωn(t1+tp)[sin(ωdt1+ωdtp+)]x2=Xeξωn(t1+tp)[sin(ωdt1+ωd(2πωd)+)]x2=Xeξωn(t1+tp)[sin(ωdt1+2π+)]x2=Xeξωn(t1+tp)[sin{2π+(ωdt1+)}]x2=Xeξωn(t1+tp)[sin(ωdt1+)]\begin{aligned} x_2 &= Xe^{-\xi \omega_n t_2} [sin(\omega_d t_2 + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ \omega_d (t_1 + t_p) + \varnothing \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d t_p + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d \bigg({2 \pi \over \omega_d}\bigg) + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + 2 \pi + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ 2 \pi + (\omega_d t_1 + \varnothing) \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] \\ \end{aligned}

Now,
Taking ratio, we get;

x1x2=Xeξωnt1[sin(ωdt1+)]Xeξωn(t1+tp)[sin(ωdt1+)]x1x2=eξωn(t1t1tp)x1x2=eξωntp\begin{aligned} {x_1 \over x_2} &= {Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)] \over Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] }\\ \therefore \quad \quad {x_1 \over x_2} &= e^{-\xi \omega_n (t_1-t_1-t_p)} \\ \therefore \quad \quad {x_1 \over x_2} &= e^{\xi \omega_n t_p} \end{aligned}

Now,
The logarithmic decrement is given by;

δ=loge(x1x2)δ=loge(eξωntp)δ=ξωntpδ=ξωn2π(1ξ2) ωnδ=2πξ(1ξ2)\begin{aligned} \delta &= log_e \bigg({x_1 \over x_2}\bigg) \\ \therefore \quad \delta &= log_e (e^{\xi \omega_n t_p}) \\ \therefore \quad \delta &= \xi \omega_n t_p \\ \therefore \quad \delta &= \xi \omega_n {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \\ \therefore \quad \delta &= {2 \pi \xi\over \big(\sqrt {1-\xi^2}\big)} \end{aligned}

The logarithmic decrement can also be determined as follows;

δ=loge(x0x1)=loge(x1x2)=loge(x2x3)==loge(xn1xn)Adding upto ’n’ termsnδ=loge(x0x1)+loge(x1x2)+loge(x2x3)++loge(xn1xn)  nδ=loge(x0x1 .x1x2 .x2x3 . .xn1xn)Ornδ=loge(x0xn)δ=1nloge(x0xn)\begin{aligned} \delta &= log_e \bigg({x_0 \over x_1}\bigg)= log_e \bigg({x_1 \over x_2}\bigg)= log_e \bigg({x_2 \over x_3}\bigg) = \dotso = log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \text {Adding upto 'n' terms}\\ n\delta &= log_e \bigg({x_0 \over x_1} \bigg) + log_e \bigg({x_1 \over x_2}\bigg) + log_e \bigg({x_2 \over x_3}\bigg) + \dotso + log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \therefore \ \ n\delta &= log_e \bigg({x_0 \over x_1} \ . {x_1 \over x_2} \ . {x_2 \over x_3} \ . \dots \ . {x_{n-1} \over x_n}\bigg) \\ \text {Or} \qquad \\ n\delta &= log_e \bigg({x_0 \over x_n}\bigg) \\ \therefore \quad \delta &= {1 \over n} log_e \bigg({x_0 \over x_n}\bigg) \end{aligned}

where,
x0x_0 = amplitude at the starting position
xnx_n = amplitude after ‘n’ cycles

fig_logarithmic-decrement fig-1: displacement V/s time curve for under damped system

Logarithmic decrement

Logarithmic decrement is defined as the natural logarithm of the ratio of successive amplitude on the same side of mean position.

The rate of decay in the amplitudes of under-damped system is measured by the parameter known as logarithmic decrement.

Rate of decay in amplitudes depends on the amount of damping present in the system. So if the damping is more, then the rate of decay will also be more.

Let A and B are the two points on the successive cycles which shows maximum deflection as shown in figure.

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The periodic time:

tp=t2t1=2πωd=2π(1ξ2) ωn\begin{aligned} t_p &= t_2 − t_1 \\ &= {2 \pi \over \omega_d} \\ &= {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \end{aligned}

The amplitude at time t1t_1 and t2t_2 are:

x1=Xeξωnt1[sin(ωdt1+)]x_1 = Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)]

And,

x2=Xeξωnt2[sin(ωdt2+)]x2=Xeξωn(t1+tp)[sin{ωd(t1+tp)+}]x2=Xeξωn(t1+tp)[sin(ωdt1+ωdtp+)]x2=Xeξωn(t1+tp)[sin(ωdt1+ωd(2πωd)+)]x2=Xeξωn(t1+tp)[sin(ωdt1+2π+)]x2=Xeξωn(t1+tp)[sin{2π+(ωdt1+)}]x2=Xeξωn(t1+tp)[sin(ωdt1+)]\begin{aligned} x_2 &= Xe^{-\xi \omega_n t_2} [sin(\omega_d t_2 + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ \omega_d (t_1 + t_p) + \varnothing \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d t_p + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d \bigg({2 \pi \over \omega_d}\bigg) + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + 2 \pi + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ 2 \pi + (\omega_d t_1 + \varnothing) \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] \\ \end{aligned}

Now,
Taking ratio, we get;

x1x2=Xeξωnt1[sin(ωdt1+)]Xeξωn(t1+tp)[sin(ωdt1+)]x1x2=eξωn(t1t1tp)x1x2=eξωntp\begin{aligned} {x_1 \over x_2} &= {Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)] \over Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] }\\ \therefore \quad \quad {x_1 \over x_2} &= e^{-\xi \omega_n (t_1-t_1-t_p)} \\ \therefore \quad \quad {x_1 \over x_2} &= e^{\xi \omega_n t_p} \end{aligned}

Now,
The logarithmic decrement is given by;

δ=loge(x1x2)δ=loge(eξωntp)δ=ξωntpδ=ξωn2π(1ξ2) ωnδ=2πξ(1ξ2)\begin{aligned} \delta &= log_e \bigg({x_1 \over x_2}\bigg) \\ \therefore \quad \delta &= log_e (e^{\xi \omega_n t_p}) \\ \therefore \quad \delta &= \xi \omega_n t_p \\ \therefore \quad \delta &= \xi \omega_n {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \\ \therefore \quad \delta &= {2 \pi \xi\over \big(\sqrt {1-\xi^2}\big)} \end{aligned}

The logarithmic decrement can also be determined as follows;

δ=loge(x0x1)=loge(x1x2)=loge(x2x3)==loge(xn1xn)Adding upto ’n’ termsnδ=loge(x0x1)+loge(x1x2)+loge(x2x3)++loge(xn1xn)  nδ=loge(x0x1 .x1x2 .x2x3 . .xn1xn)Ornδ=loge(x0xn)δ=1nloge(x0xn)\begin{aligned} \delta &= log_e \bigg({x_0 \over x_1}\bigg)= log_e \bigg({x_1 \over x_2}\bigg)= log_e \bigg({x_2 \over x_3}\bigg) = \dotso = log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \text {Adding upto 'n' terms}\\ n\delta &= log_e \bigg({x_0 \over x_1} \bigg) + log_e \bigg({x_1 \over x_2}\bigg) + log_e \bigg({x_2 \over x_3}\bigg) + \dotso + log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \therefore \ \ n\delta &= log_e \bigg({x_0 \over x_1} \ . {x_1 \over x_2} \ . {x_2 \over x_3} \ . \dots \ . {x_{n-1} \over x_n}\bigg) \\ \text {Or} \qquad \\ n\delta &= log_e \bigg({x_0 \over x_n}\bigg) \\ \therefore \quad \delta &= {1 \over n} log_e \bigg({x_0 \over x_n}\bigg) \end{aligned}

where,
x0x_0 = amplitude at the starting position
xnx_n = amplitude after ‘n’ cycles

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