The frequency of A note is $4$ times that of B note. The energies of two notes are equal. The amplitude of B note as compared to that of A note will be:

double

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equal

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four times

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eight times

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Solution

The correct option is A double
$E$ is for energy, $A$ for amplitude and $f$ for frequency.
As per the problem $E_{A} = E_{B}$
Hence, $f_{A} \times A_{A}^{2} = f_{B} \times A_{B}^{2}$
$4 f_{B} \times A_{A}^{2} = f_{B} \times A_{B}^{2}$
Hence, $2 A_{A} = A_{B}$

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The frequency of A note is 4 times that of B note. The energies of two notes are equal. The amplitude of B note as compared to that of A note will be:

The correct option is A doubleE is for energy, A for amplitude and f for frequency.As per the problem EA=EBHence, fA×A2A=fB×A2B 4fB×A2A=fB×A2BHence, 2AA=AB

The frequency of A note is $4$ times that of B note. The energies of two notes are equal. The amplitude of B note as compared to that of A note will be:

The correct option is A double
$E$ is for energy, $A$ for amplitude and $f$ for frequency.
As per the problem $E_{A} = E_{B}$
Hence, $f_{A} \times A_{A}^{2} = f_{B} \times A_{B}^{2}$
$4 f_{B} \times A_{A}^{2} = f_{B} \times A_{B}^{2}$
Hence, $2 A_{A} = A_{B}$