Normal Modes of Stretched String – Rankers Physics

Standing Wave in String and Organ Pipe: Practice Problem & Solution

Assertion (A): In \(n^{\text{th}}\) normal mode of a stretched string, there are \(n\) antinodes and \((n+1)\) nodes. Reason (R): The ends of string are nodes, so the number of nodes is one more than the number of antinodes.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Standing Wave in String and Organ Pipe. Understanding the underlying formula is key to arriving at the correct answer below:

Assertion (A) is true. For a stretched string fixed at both ends, the \(n^{\text{th}}\) normal mode indeed has \(n\) antinodes and \(n+1\) nodes.

Reason (R) is true and correctly explains this: the fixed ends are always nodes, leading to one more node than antinode.

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