Question

If the volume of a room is doubled and the total absorption is halved, the reverberation time will:

• A. Remain unchanged
• B. Be doubled
• C. Become four times
• D. Be halved

Hint:

Double the size ($ \times 2$) and half the sponges ($ \times 2$) = 4 times the echo time.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Reverberation time (\(T\)) is the time required for sound to decay by 60 decibels. It is governed by Sabine’s Formula, which relates the volume of the space to its sound-absorbing properties.
Step 2: Key Formula or Approach:
Sabine’s Formula: \( T = \frac{0.16V}{A} \) Where \(V\) is the volume and \(A\) is the total absorption.
Step 3: Detailed Explanation:
Let the initial reverberation time be \(T_1 = \frac{0.16V}{A}\). According to the problem: 1. New volume \(V' = 2V\) 2. New absorption \(A' = \frac{A}{2}\) Substitute these into the formula for the new reverberation time \(T_2\): \[ T_2 = \frac{0.16(2V)}{A/2} = \frac{0.16 \times 2V \times 2}{A} \] \[ T_2 = 4 \times \left( \frac{0.16V}{A} \right) = 4T_1 \]
Step 4: Final Answer:
The reverberation time will become four times its original value.