Question

3. A string 80 cm long has fundamental frequency of 300 vib/sec. In the length of the string is changed
to 40 cm without changing the tension, the fundamental frequency will be in vib/sec:
B. 150unde root 2
C. 300
D. 300 under root 2
A. 150
​

Answer 1

Explanation:

fundamental frequency of vibration = f , L = legth of the string = 80 cm = 0.80 m 

T ension in the string = T , m = mass per unit length of the string 

f = (1 / 2L) * √ (T / m) 

T and m remaining same, f propotional to (1/L)

(f 1) / (f 2) = (L2) / (L1) ; f1 = 300 Hz, L1 = 80 cm, L2 = 40 cm, f2 = to be found out 

(f 2) / (f 1) = (L1) / (L2) ==> (f 2) = (L1 / L2) * (f 1) = (80/40)*(300) Hz = 600 Hz

Answer 2

GiveN :

• L = legth of the string = 80 cm = 0.80 m
• Fundamental frequency = 300 vib/sec
• Length of string is changed to = 40 cm

We know the required formula for finding out the fundamental frequency.

$\longrightarrow\sf f = \dfrac{1}{2L} \times \sqrt{\dfrac{T}{m}}$

In this case the value of T & m would be constant. Hence we can conclude that,

$\because \: { \boxed{ \sf f \propto \dfrac{1}{2L}}}$

The new formula obtained would be :

$\longrightarrow\sf \dfrac{f1}{f2}=\dfrac{L2}{L1}$

Now, we've the values :

• f1 = 300 Hz

• f2 = unknown

• L1 = 80 cm

• L2 = 40 cm

Substituting the values :

$\longrightarrow\sf \dfrac{f1}{f2}=\dfrac{L2}{L1}$

$\longrightarrow\sf f2 =\dfrac{L2}{L1}\times f1$

$\longrightarrow\sf f2=\dfrac{80}{40}\times300$

$\longrightarrow \large{ \boxed{\frak{f2= 600 \: Hz}}}$

$\therefore$ The fundamental frequency will be F2 = 600 Hz.[Correction in your options]