Question

17. A small wave is sent along a stretched
string with a velocity of 150 ms - If the
string is 50 cm long, density is 8 gm/c.c and
the area of cross section is 6x10 cm.
Calculate the tension applied on the string.
​

Answer 1

Given:

A small wave is sent along a stretched string with a velocity of 150 m/s. The string is 50 cm long, density is 8 gm/c.c and the area of cross section is 6x10^(-2) cm².

To find:

Tension on the string?

Calculation:

We know that velocity of a transverse wave on a string is given as:

$\sf \: v = \sqrt{ \dfrac{T}{ \mu} }$

• 'T' is tension and $\mu$ is mass per unit length.

Putting the necessary values in SI UNIT:

$\sf \implies\: 150= \sqrt{ \dfrac{T}{( \frac{m}{l}) } }$

$\sf \implies\: 150= \sqrt{ \dfrac{T}{( \frac{volume \times density}{l}) } }$

$\sf \implies\: 150= \sqrt{ \dfrac{T}{( \frac{6 \times {10}^{ - 6} \times 0.5 \times 8000}{0.5}) } }$

$\sf \implies\: 150= \sqrt{ \dfrac{T}{( 6 \times {10}^{ - 6} \times 8000) } }$

$\sf \implies\: 150= \sqrt{ \dfrac{T}{48 \times {10}^{ - 3} } }$

$\sf \implies\: 22500= \dfrac{T}{48 \times {10}^{ - 3} }$

$\sf \implies\: T = 22500 \times (48 \times {10}^{ - 3} )$

$\sf \implies\: T =10800 \times {10}^{ - 1}$

$\sf \implies\: T =1080 \: N$

So, tension is 1080 N.

Answer 2

Answer:

$Given:</p><p></p><p>A small wave is sent along a stretched string with a velocity of 150 m/s. The string is 50 cm long, density is 8 gm/c.c and the area of cross section is 6x10^(-2) cm².</p><p></p><p>To find:</p><p></p><p>Tension on the string?</p><p></p><p>Calculation:</p><p></p><p>We know that velocity of a transverse wave on a string is given as:</p><p></p><p>\sf \: v = \sqrt{ \dfrac{T}{ \mu} }v=μT</p><p></p><p>'T' is tension and \muμ is mass per unit length.</p><p></p><p>Putting the necessary values in SI UNIT:</p><p></p><p>\sf \implies\: 150= \sqrt{ \dfrac{T}{( \frac{m}{l}) } }⟹150=(lm)T</p><p></p><p>\sf \implies\: 150= \sqrt{ \dfrac{T}{( \frac{volume \times density}{l}) } }⟹150=(lvolume×density)T</p><p></p><p>\sf \implies\: 150= \sqrt{ \dfrac{T}{( \frac{6 \times {10}^{ - 6} \times 0.5 \times 8000}{0.5}) } }⟹150=(0.56×10−6×0.5×8000)T</p><p></p><p>\sf \implies\: 150= \sqrt{ \dfrac{T}{( 6 \times {10}^{ - 6} \times 8000) } }⟹150=(6×10−6×8000)T</p><p></p><p>\sf \implies\: 150= \sqrt{ \dfrac{T}{48 \times {10}^{ - 3} } }⟹150=48×10−3T</p><p></p><p>\sf \implies\: 22500= \dfrac{T}{48 \times {10}^{ - 3} }⟹22500=48×10−3T</p><p></p><p>\sf \implies\: T = 22500 \times (48 \times {10}^{ - 3} )⟹T=22500×(48×10−3)</p><p></p><p>\sf \implies\: T =10800 \times {10}^{ - 1}⟹T=10800×10−1</p><p></p><p>\sf \implies\: T =1080 \: N⟹T=1080N</p><p></p><p>So, tension is 1080 N.</p><p></p><p>$