The student finds that a standing wave with one antinode is formed on the string when the frequency of the oscillator is f0. The student then moves the oscillator to shorten the horizontal segment of string to a fourth of its original length. At what frequency will a standing wave with one antinode now be formed on the string?

Explanation:

A standing wave with one antinode in a string means fundamental mode of vibration  being formed . In this case

wave length = 2 x length of string

λ = 2L

frequency = velocity of sound / wavelength

= V / 2L

f₀ = V / 2L

new length

l = L / 4

When one antinode now be formed on the string

again fundamental mode of vibration

wave length = 2 x length of string

λ₁ = 2l  , λ₁  is new wavelength , l is new length .

frequency = velocity of sound / wavelength

F₀ = V / 2l , F₀ is new frequency,

F₀ = 4V / 2L

= 4 ( V / 2L )

= 4 x f₀

required frequency = 4f₀

four times the previous frequency will be required.

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i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.