Question

A quartz crystal of thickness 1 mm is vibrating at resonance. Calculate the
fundamental frequency. Given Y for quartz =
$7.9 \times {10}^{10}$
Nm-2 and p for quartz
= 2650 kg m-3.​

Answer 1

Answer:

Fundamental frequency = 2.72947 × 10⁶ Hz

Explanation:

Given:

• Thickness of the quartz crystal = 1 mm = 10⁻³m
• Y of quartz = 7.9 × 10¹⁰ N/m²
• ρ of quartz = 2650 kg/m³

To Find:

• The fundamental frequency

Solution:

Here given that the quartz crystal is vibrating at resonance.

The fundamental frequency is given by,

$\sf Fundamental\:frequency=\dfrac{v}{2t}$

where v is the velocity and t is the thickness.

We know that,

$\sf v=\sqrt{\dfrac{Y}{\rho} }$

where Y is the Young's modulus and ρ is the density of the solid.

Substituting the data we get,

$\sf v =\sqrt{\dfrac{7.9\times 10^{10} }{2650} }$

$\sf v = \sqrt{2.98\times 10^{7} }$

$\sf v = 5458.94\:m/s$

Substitute the value of v in the above equation,

$\sf Fundamental\:frequency=\dfrac{5458.94}{2\times 10^{-3} }$

$\sf \implies 2729.47 \times 10^{3} \: Hz$

$\sf \implies 2.72947\times 10^{6} \:Hz$

Hence the fundamental frequency is 2.72947 × 10⁶ Hz.

Answer 2

$\: \: \: \: \:{\large{\bold{\sf{\underline{Correct \; Question}}}}}$

A quartz crystal of thickness 1 mm is vibrating at resonance. Calculate the

fundamental frequency. Given Y for quartz = 7.9 × 10¹⁰ N/m² and ρ for quartz = 2650 kg/m³

${\large{\bold{\sf{\underline{Given \; that}}}}}$

${\bullet}$ A quartz crystal of thickness 1 mm

${\bullet}$ Quartz's Y = 7.9 × 10¹⁰ N/m²

${\bullet}$ Quartz's ρ = 2650 kg/m³

${\large{\bold{\sf{\underline{To \; calculate}}}}}$

${\bullet}$ Fundamental frequency

${\large{\bold{\sf{\underline{Solution}}}}}$

${\bullet}$ Fundamental frequency = 2.72947 × 10⁶ Hz

${\large{\bold{\sf{\underline{Using \; concept}}}}}$

${\bullet}$ Fundamental frequency

${\bullet}$ Velocity ( Fundamental frequency )

${\large{\bold{\sf{\underline{Using \; rule}}}}}$

${\bullet}$ $\; \: \; \; \; \;{\boxed{\boxed{\rm{Fundamental \; frequency = \dfrac{v}{2t}}}}}$

Where,

• v denotes velocity.

• t denotes thickness.

${\bullet}$ $\; \; \; \; \; \;{\boxed{\boxed{\rm{v=\sqrt \dfrac{Y}{\rho}}}}}$

Where,

• Y denotes Young's modulus

• ρ denotes Solid density

${\large{\bold{\sf{\underline{Full \; Solution}}}}}$

~ Using formula let's put the values and find the fundamental frequency !

$\; \; \; \; \; \;{\bf{\longmapsto v=\sqrt \dfrac{Y}{\rho}}}$

$\; \: \; \; \; \;{\bf{\longmapsto \sqrt \dfrac{7.9\times 10^{10}}{2650}}}$

$\; \; \; \; \; \;{\bf{\longmapsto v = \sqrt 2.98\times 10^{7}}}$

$\; \: \; \; \; \;{\bf{\longmapsto v = 5458.94 \; m/s}}$

~ Now let's substitute the value of v as 5458.94

$\; \: \; \; \; \;{\bf{\longmapsto Fundamental \: frequency = \dfrac{5458.94}{2\times 10^{-3}}}}$

$\; \: \; \; \; \;{\bf{\longmapsto Fundamental \: frequency = 2729.47 \times 10^{3} \: Hz}}$

$\; \: \; \; \; \;{\bf{\longmapsto Fundamental \: frequency =2.72947\times 10^{6} \:Hz}}$

${\pink{\frak{Henceforth, \: 2.72947 \times 10^{6} \: Hz \: is \: the \: fundamental \: frequency}}}$