Question:
A steel wire of length \(1.25\ \text{m}\) is stretched between two rigid supports. The tension in the wire produces an elastic strain of \(0.14\%\). The fundamental frequency of the wire is
\[
(\text{Density and Young's modulus of steel are }7.7\times10^3\ \text{kg m}^{-3}\text{ and }2.2\times10^{11}\ \text{N m}^{-2}\text{ respectively})
\]
A steel wire of length \(1.25\ \text{m}\) is stretched between two rigid supports. The tension in the wire produces an elastic strain of \(0.14\%\). The fundamental frequency of the wire is
\[ (\text{Density and Young's modulus of steel are }7.7\times10^3\ \text{kg m}^{-3}\text{ and }2.2\times10^{11}\ \text{N m}^{-2}\text{ respectively}) \]
\[ (\text{Density and Young's modulus of steel are }7.7\times10^3\ \text{kg m}^{-3}\text{ and }2.2\times10^{11}\ \text{N m}^{-2}\text{ respectively}) \]
Show Hint
For a stretched wire,
\[
f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}.
\]
If strain is given, use
\[
\frac{T}{\mu}=\frac{Y\times \text{strain}}{\rho}.
\]
This eliminates the need to know the cross-sectional area of the wire.
Updated On: Jun 26, 2026
- \(20\ \text{Hz}\)
- \(40\ \text{Hz}\)
- \(80\ \text{Hz}\)
- \(160\ \text{Hz}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Write the formula for fundamental frequency.
For a stretched wire fixed at both ends, the fundamental frequency is \[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}, \] where \(L\) is the length of the wire, \(T\) is the tension, and \(\mu\) is the mass per unit length.
Step 2: Relate tension to Young's modulus and strain.
Young's modulus is \[ Y=\frac{\text{stress}}{\text{strain}}. \] So, \[ \text{stress}=Y\times \text{strain}. \] Also, \[ \text{stress}=\frac{T}{A}, \] where \(A\) is the area of cross-section.
Thus, \[ \frac{T}{A}=Y\times \text{strain}. \] So, \[ T=AY\times \text{strain}. \] Mass per unit length is \[ \mu=\rho A. \] Therefore, \[ \frac{T}{\mu} = \frac{AY(\text{strain})}{\rho A}. \] Canceling \(A\), \[ \frac{T}{\mu} = \frac{Y(\text{strain})}{\rho}. \]
Step 3: Substitute the given values.
Given, \[ L=1.25\ \text{m}, \] \[ Y=2.2\times10^{11}\ \text{N m}^{-2}, \] \[ \rho=7.7\times10^3\ \text{kg m}^{-3}. \] Elastic strain is \[ 0.14\%=\frac{0.14}{100}=0.0014. \] Therefore, \[ f=\frac{1}{2L}\sqrt{\frac{Y(\text{strain})}{\rho}}. \] \[ f=\frac{1}{2(1.25)} \sqrt{\frac{(2.2\times10^{11})(0.0014)}{7.7\times10^3}}. \]
Step 4: Simplify the expression.
First, \[ (2.2\times10^{11})(0.0014) = 3.08\times10^8. \] Now, \[ \frac{3.08\times10^8}{7.7\times10^3} = 4.0\times10^4. \] So, \[ \sqrt{\frac{Y(\text{strain})}{\rho}} = \sqrt{4.0\times10^4}. \] \[ =200\ \text{m s}^{-1}. \] Thus, \[ f=\frac{200}{2.5}. \] \[ f=80\ \text{Hz}. \]
Step 5: Final conclusion.
Therefore, the fundamental frequency of the wire is \[ \boxed{80\ \text{Hz}} \] Hence, the correct option is \[ \boxed{(3)} \]
For a stretched wire fixed at both ends, the fundamental frequency is \[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}, \] where \(L\) is the length of the wire, \(T\) is the tension, and \(\mu\) is the mass per unit length.
Step 2: Relate tension to Young's modulus and strain.
Young's modulus is \[ Y=\frac{\text{stress}}{\text{strain}}. \] So, \[ \text{stress}=Y\times \text{strain}. \] Also, \[ \text{stress}=\frac{T}{A}, \] where \(A\) is the area of cross-section.
Thus, \[ \frac{T}{A}=Y\times \text{strain}. \] So, \[ T=AY\times \text{strain}. \] Mass per unit length is \[ \mu=\rho A. \] Therefore, \[ \frac{T}{\mu} = \frac{AY(\text{strain})}{\rho A}. \] Canceling \(A\), \[ \frac{T}{\mu} = \frac{Y(\text{strain})}{\rho}. \]
Step 3: Substitute the given values.
Given, \[ L=1.25\ \text{m}, \] \[ Y=2.2\times10^{11}\ \text{N m}^{-2}, \] \[ \rho=7.7\times10^3\ \text{kg m}^{-3}. \] Elastic strain is \[ 0.14\%=\frac{0.14}{100}=0.0014. \] Therefore, \[ f=\frac{1}{2L}\sqrt{\frac{Y(\text{strain})}{\rho}}. \] \[ f=\frac{1}{2(1.25)} \sqrt{\frac{(2.2\times10^{11})(0.0014)}{7.7\times10^3}}. \]
Step 4: Simplify the expression.
First, \[ (2.2\times10^{11})(0.0014) = 3.08\times10^8. \] Now, \[ \frac{3.08\times10^8}{7.7\times10^3} = 4.0\times10^4. \] So, \[ \sqrt{\frac{Y(\text{strain})}{\rho}} = \sqrt{4.0\times10^4}. \] \[ =200\ \text{m s}^{-1}. \] Thus, \[ f=\frac{200}{2.5}. \] \[ f=80\ \text{Hz}. \]
Step 5: Final conclusion.
Therefore, the fundamental frequency of the wire is \[ \boxed{80\ \text{Hz}} \] Hence, the correct option is \[ \boxed{(3)} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Wave Motion Questions
- If the speed of a wave doubles as it passes from shallow water into deeper water, its wavelength will be
- The frequency of a wave is 500 Hz and its wavelength is 0.66 m. What is the speed of the wave?
- The ratio of radii of two wires is 1:2 and the density of their materials are in the ratio 1:4. If same tension is applied to both the wires then the ratio of the speed of transverse waves produced in them is:
- A wave is represented by \( y = 0.1 \sin(200t - 10x) \). Find wave velocity.
- A source of sound moves towards a stationary observer with speed \( 10 \text{ m/s} \). Speed of sound = \( 340 \text{ m/s} \). If original frequency is \( 340 \text{ Hz} \), apparent frequency is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions