Question 452: Waves

Question

When tension $$T$$ is applied to a sonometer wire of length $$I$$, it vibrates with the fundamental frequency $$n$$. Keeping the experimental setup same, when the tension is increased by 8 N, the fundamental frequency becomes three times the earlier fundamental frequency $$n$$. The initial tension applied to the wire (in newton) was

Accepted Answer

When a tension $$T$$ is applied to a sonometer wire of length $$L$$, it vibrates with a fundamental frequency $$n$$. If the tension is increased by 8 N, the fundamental frequency becomes three times the initial frequency. We need to find the initial tension applied to the wire.
The formula for the fundamental frequency of a vibrating wire is given by:
$ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \quad 	ext{.... (i)} $
Where $$\mu$$ is the linear density of the wire.
For the increased tension, the fundamental frequency becomes:
$ 3n = \frac{1}{2L} \sqrt{\frac{T+8}{\mu}} \quad 	ext{... (ii)} $
To find the relationship between the initial and final tensions, we divide equation (ii) by equation (i):
$ \frac{3n}{n} = \frac{\frac{1}{2L} \sqrt{\frac{T+8}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T}{\mu}}} $
This simplifies to:
$  3 = \sqrt{\frac{T+8}{T}}  $
Squaring both sides gives:
$  9T = T + 8  $
Solving for $T$:
$ 8T = 8 \quad \Rightarrow \quad T = 1 	ext{ N} $
Thus, the initial tension applied to the wire was $1 	ext{ N}$.

Answer

2.0

Answer

2.5

Answer

0.5

Answer

1.0

Waves

Choose the correct answer: