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If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:
- a)√2 :1
- b)1 : √2
- c)1 : 2
- d)1 : 1
Correct answer is option 'B'. Can you explain this answer?
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If the initial tension on a stretched string is doubled, then the rati...
Explanation:
Initial and Final Speeds of a Transverse Wave:
When the initial tension on a stretched string is doubled, the speed of a transverse wave along the string changes. The speed of a transverse wave on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where:
- \( v \) = speed of the wave
- \( T \) = tension in the string
- \( \mu \) = mass per unit length of the string
Ratio of Initial and Final Speeds:
Let's denote the initial tension as \( T_1 \) and the final tension as \( T_2 \).
According to the question, the initial tension is doubled, so \( T_2 = 2T_1 \).
The speed of the wave is directly proportional to the square root of tension. Therefore, the ratio of initial and final speeds can be calculated as:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{\mu}} : \sqrt{\frac{2T_1}{\mu}} = \sqrt{\frac{T_1}{\mu}} : \sqrt{\frac{T_1}{\mu}}\sqrt{2} = 1 : \sqrt{2} \]
So, the correct answer is option B which is 1 : \( \sqrt{2} \).
Initial and Final Speeds of a Transverse Wave:
When the initial tension on a stretched string is doubled, the speed of a transverse wave along the string changes. The speed of a transverse wave on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where:
- \( v \) = speed of the wave
- \( T \) = tension in the string
- \( \mu \) = mass per unit length of the string
Ratio of Initial and Final Speeds:
Let's denote the initial tension as \( T_1 \) and the final tension as \( T_2 \).
According to the question, the initial tension is doubled, so \( T_2 = 2T_1 \).
The speed of the wave is directly proportional to the square root of tension. Therefore, the ratio of initial and final speeds can be calculated as:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{\mu}} : \sqrt{\frac{2T_1}{\mu}} = \sqrt{\frac{T_1}{\mu}} : \sqrt{\frac{T_1}{\mu}}\sqrt{2} = 1 : \sqrt{2} \]
So, the correct answer is option B which is 1 : \( \sqrt{2} \).
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If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:a)√2 :1b)1 : √2c)1 : 2d)1 : 1Correct answer is option 'B'. Can you explain this answer?
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