Tension

To determine the ratio of the new tension to the original tension for the vibrating string, we should consider the fundamental frequency formula for a stretched string:

$f%20%3D%20%5Cdfrac%7B1%7D%7B2L%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7B%5Cmu%7D%7D$

Where:

$f$ is the frequency of vibration.
$L$ is the length of the string.
$T$ is the tension in the string.
$%5Cmu$ is the linear mass density of the string, assumed constant in both cases.

Initially, the frequency $f_1$ is 200 Hz, and the length is $L$ .

When the string length is doubled to $2L$ and the tension changes, the frequency becomes $f_2%20%3D%20300$ Hz.

Using the formula for frequency in both scenarios, we have:

$f_1%20%3D%20%5Cdfrac%7B1%7D%7B2L%7D%5Csqrt%7B%5Cdfrac%7BT_1%7D%7B%5Cmu%7D%7D%20%3D%20200$

$f_2%20%3D%20%5Cdfrac%7B1%7D%7B4L%7D%5Csqrt%7B%5Cdfrac%7BT_2%7D%7B%5Cmu%7D%7D%20%3D%20300$

Divide the second equation by the first equation:

$%5Cdfrac%7Bf_2%7D%7Bf_1%7D%20%3D%20%5Cdfrac%7B%5Cdfrac%7B1%7D%7B4L%7D%5Csqrt%7B%5Cdfrac%7BT_2%7D%7B%5Cmu%7D%7D%7D%7B%5Cdfrac%7B1%7D%7B2L%7D%5Csqrt%7B%5Cdfrac%7BT_1%7D%7B%5Cmu%7D%7D%7D$

After simplification:

$%5Cdfrac%7B300%7D%7B200%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B%5Cdfrac%7BT_2%7D%7BT_1%7D%7D$

By solving the frequency ratio, we find:

$%5Cdfrac%7B3%7D%7B2%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B%5Cdfrac%7BT_2%7D%7BT_1%7D%7D$

Now, solve for $%5Cdfrac%7BT_2%7D%7BT_1%7D$ :

$%5Csqrt%7B%5Cdfrac%7BT_2%7D%7BT_1%7D%7D%20%3D%203$

Squaring both sides yields:

$%5Cdfrac%7BT_2%7D%7BT_1%7D%20%3D%209$

Hence, the ratio of the new tension to the original tension is 9:1.

The problem involves a particle of mass $m$ connected to a string of length $l$ , which moves in a circular path on a smooth horizontal table. The force of tension $T$ in the string acts as the centripetal force, which keeps the particle moving in a circle at speed $v$ . The centripetal force $F_c$ required to keep an object moving in a circle of radius $r$ at speed $v$ is given by:

$F_c%20%3D%20%5Cfrac%7Bm%20v%5E2%7D%7Br%7D$

Here, the radius of the circle is the length of the string $l$ , so the expression becomes:

$F_c%20%3D%20%5Cfrac%7Bm%20v%5E2%7D%7Bl%7D$

Since the only force acting towards the center is the tension in the string, we equate it to the centripetal force:

$T%20%3D%20%5Cfrac%7Bm%20v%5E2%7D%7Bl%7D$

Thus, the tension $T$ itself serves as the required centripetal force.
Therefore, the net force on the particle directed towards the center is: $T$

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Tension

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs