Question
Question: A wire of length l having tension T and radius r vibrates with natural frequency f. Another wire of ...
A wire of length l having tension T and radius r vibrates with natural frequency f. Another wire of same metal with length 2l having tension 2T and radius 2r will vibrate with natural frequency
A. f.
B. 2f.
C. 22f.
D. 2f.
Solution
To solve this question, we have to remember that the natural frequency of a vibrating string can be given as: 2L1mT, where L is the length of the string, r is the radius, T is the tension of the string and m is the mass per unit length of the string.
Complete answer:
We have given that, when
Length of wire = L,
Tension = T and
Radius = r, the wire will vibrate with natural frequency f.
As we know that,
Natural frequency = 2L1mT
Putting these values in this formula, we will get
⇒f=2L1mT …….. (i)
According to the question, if
Length of wire become 2L, i.e. L′=2L
Tension become 2T, i.e. T′=2T and
Radius become 2r, i.e. r′=2r
We have to find the natural frequency with these conditions, f′ on the same wire, i.e. the mass per unit length (m) will remain constant.
Putting these values in the formula of natural frequency,
⇒f′=2L′1mT′
Putting the value of L′and T′, we will get
⇒f′=2×2L1m2T
⇒f′=2×2L2mT
⇒f′=21(2L1mT)
From equation (i), putting the value
⇒f′=2f
Hence, we can say that the natural frequency of the same string will become 2f
So, the correct answer is “Option D”.
Note:
Whenever we are asked these types of questions, we will use the basic concept of the fundamental frequency of a stretched string. When the wave relationship is applied to a stretched string, it is seen that resonant standing wave modes are produced. The lowest frequency mode for a stretched string is called the fundamental frequency.