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Question

In a sonometer experiment, the string of length L under tension vibrates in a second overtone between two bridges. Where is the amplitude of vibration maximum at?


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Solution

In a sonometer experiment the string of length L under class 11 physics CBSE

Step 1:Given

Given that the length of the string is L.

Under tension vibrates in a second overtone between two bridges.

Therefore, it has two nodes and three antinodes, so we will get three maxima.

n=3

Let, the wavelength of the string be λ.

Step 2:: Derive the wavelength

The length of the string in terms of wavelength can be given by

L=nλ2

Upon substituting the value of n we get,

L=3λ2

λ=2L3

Step 3: Calculate the three maxima

Hence, we can say that the first maximum is appearing at wavelength,

λ4=14×23L

=L6

Similarly, second amplitude maxima will be at

λ4+λ2=34λ

=34×23L

=L2

And, third amplitude maxima will be at

34λ+λ2=54λ

=54×23L

=5L6

Step 4: Estimate the highest maximum

Taking the LCM we can rewrite the maxima as,

5L6,3L6,L6

Upon comparing these three maxima we can see that

5L6>3L6>L6

Hence, the amplitude of vibration maximum will be at 5L6.


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