Question

The diameter of a stretched string is increased 2% keeping the other parameters constant then the velocity

Answer 1

Given : The diameter of a stretched string is increased 2% keeping the other parameters constant.

To find : The percentage change in velocity.

solution : velocity of transverse wave in a string is given by...

v = √{T/(m/l)}

where T is tension in string , m is mass of string and l is length of string.

if ρ is density of material of string and d is the radius of string

then, mass of string, m = πd²/4lρ

then, v = √{4T/πd²ρ}

as diameter of a stretched string is increased by 2%. (> 10%) we should use error measurements concept.

here, v² = 4T/πd²ρ

2∆v/v = ∆T/T + 2∆d/d

as T remains constant. so, ∆T/T = 0

so, 2∆v/v = 2∆d/d

⇒∆v/v × 100 = ∆d/d × 100

⇒% change in velocity = % change in diameter

⇒% change in velocity = 2%

therefore velocity is also increased by 2%.