Find the natural frequencies and mode shapes of a taut string supported subject to different conditions

Question No 03
Find the natural frequencies and mode shapes of a taut string supported
subject to different conditions as shown in the figure given below. The
initial conditions are given as:
𝝎(𝒙, 𝒕 = 𝟎) = 𝝎𝟎 (𝒙)
𝝏𝝎
(𝒙, 𝒕 = 𝟎)𝝎̇(𝒙)
𝝏𝒙
3. Both ends attached with masses.




Solution: -
The governing equation is
𝜕2 𝜔 𝜕2 𝜔
𝑐2 = (1)
𝜕𝑥 2 𝜕𝑡 2

Boundary conditions
𝜕𝜔 𝜕2 𝜔
𝑃 (0, 𝑡 ) = 𝑚1 (0, 𝑡) (2)
𝜕𝑥 𝜕𝑡 2

𝜕𝜔 𝜕2 𝜔
𝑃 (𝑙, 𝑡 ) = −𝑚2 (𝑙, 𝑡) (3)
𝜕𝑥 𝜕𝑡 2

Initial conditions
𝜔(𝑥, 𝑡 = 0) = 𝜔0 (𝑥) (i)
𝜕𝜔
(𝑥, 𝑡 = 0) = 𝜔̇ (𝑥) (ii)
𝜕𝑥

We use separation of variable method to find solution. For this we let:
𝜔(𝑥, 𝑡 ) = 𝑊 (𝑥)𝑇(𝑡) (4)
Substituting equation (4) in equation (1)
𝑐 2 𝑑2 𝑊 1 𝑑2 𝑇
=
𝑊 𝑑𝑥 2 𝑇 𝑑𝑡 2
Since LHS is function of 𝑥 only and RHS is the function of 𝑡 only, therefore

, 𝑐 2 𝑑2 𝑊 1 𝑑2 𝑇
= = 𝛼(𝑠𝑎𝑦)
𝑊 𝑑𝑥 2 𝑇 𝑑𝑡 2
This equation can be separated into two equation as follows:
𝑑2 𝑊 𝛼
− 𝑊=0
𝑑𝑥 2 𝑐 2
𝑑2 𝑇
− 𝛼𝑇 = 0
𝑑𝑥 2
Setting 𝛼 = −𝜔2 above equation take the form:
𝑑2 𝑊 𝜔2
+ 𝑊=0 (5)
𝑑𝑥 2 𝑐2

𝑑2 𝑇
+ 𝜔2 𝑇 = 0 (6)
𝑑𝑥 2

With boundary conditions
𝑑𝑊 𝑑2 𝑊
𝑃 (0, 𝑡 ) = 𝑚1 (0, 𝑡) (2a)
𝑑𝑥 𝑑𝑡 2

𝑑𝑊 𝑑2 𝑊
𝑃 (𝑙, 𝑡 ) = −𝑚2 (𝑙, 𝑡) (3a)
𝑑𝑥 𝑑𝑡 2

Solve equation (5)
𝑑 2 𝑊 𝜔2
+ 2𝑊=0
𝑑𝑥 2 𝑐
The characteristic equation is
𝜔2 𝑑
𝐷12 + 𝑊=0 where 𝐷1 =
𝑐2 𝑑𝑥
𝜔
⇒ 𝐷1 = ± 𝑖
𝑐
𝜔𝑥 𝜔𝑥
⇒ 𝑊 (𝑥) = 𝐴𝑐𝑜𝑠 + 𝐵𝑠𝑖𝑛 (7)
𝑐 𝑐

Now solving equation (6)
𝑑
𝐷22 + 𝑊 = 0 where 𝐷2 =
𝑑𝑥

⇒ 𝐷2 = ±𝜔𝑖
⇒ 𝑇(𝑡 ) = 𝐶𝑐𝑜𝑠𝜔𝑡 + 𝐷𝑠𝑖𝑛𝜔𝑡 (8)
Rewriting the equation (4)
For 𝑛𝑡ℎ mode of vibration. Let
𝑊 (𝑥) = 𝑊𝑛 (𝑥)
And 𝑇(𝑡 ) = 𝑇𝑛 (𝑡)