Question

what is a conical pendulum show that its time period is given by 2π√lcosΦ/g where l is length of stringΦis the angle that makes with the vertical and g is the acceleration due to gravity ​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Length of pendulum = L.
• Acceleration due to gravity = g.
• Angle of inclination = θ.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Conical Pendulum:-

Conical pendulum is defined as ,

If a mass or bob is tied to a string, and is rotated about a fixed axis Horizontally, I.e. in Horizontal circle is called a conical pendulum.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Figure,

We can Make out The Tension in the string is "F".

Now,

As we can make out,

The Tension's is resolved in to components,

It's one component is Fsinθ and Fcosθ.

The Fsinθ Component gives the Centripetal Force,

Therefore,

$\large{\boxed{\tt F = \dfrac{mv^2}{r}}}$

Substituting the values,

$\large{\tt \leadsto Fsin \theta = \dfrac{mv^2}{r}}$

$\large{\tt \leadsto Fsin \theta = \dfrac{m(r \omega)^2}{r} \: \: \: \: \rightarrow\: \: \: \: (v = r \omega)}$

Now,

$\large{\tt \leadsto Fsin\theta = \dfrac{mr^2 \omega^2}{r}}$

$\large{\tt \leadsto Fsin\theta = \dfrac{m\cancel{r^2} \omega^2}{\cancel{r}}}$

$\large{\underline{\underline{\tt Fsin \theta = m\omega^2 r}}}$

Now, From the Figure we can see,

Applying Trigonometric ratios,

$\large{\boxed{\tt Sin \theta = \dfrac{Opposite \: side}{Hypotenuse}}}$

• Opposite side = r
• Hypotenuse = L.

Substituting the values,

$\large{\tt \leadsto sin \theta = \dfrac{r}{L}}$

Substituting in the above equation, we get,

$\large{\tt \leadsto F \times \dfrac{r}{L} = m\omega^2r}$

$\large{\tt \leadsto F \times \dfrac{\cancel{r}}{L} = m\omega^2\cancel{r}}$

$\large{\tt \leadsto F \times \dfrac{1}{L} = m\omega^2}$

$\large{\tt \leadsto F = m \omega^2 L \: -----(1)}$

$\large{\boxed{\tt F = m\omega^2 L}}$

Now, From F.B.D of the mass "m",

$\large{\tt Fcos \theta = mg}$

As both are antagonistic to each other,

Now, Substituting the values from equation (1),

$\large{\tt \leadsto m \omega^2 L \times cos \theta = mg}$

$\large{\tt \leadsto \cancel{m} \omega^2 L \times cos \theta = \cancel{m}g}$

$\large{\tt \leadsto \omega^2 \times L cos \theta = g}$

$\large{\tt \leadsto \omega^2 = \dfrac{g}{L cos \theta}}$

$\large{\tt \leadsto \omega = \sqrt{\dfrac{g}{L cos \theta}}}$

As we know,

∵ω = 2π/T

Substituting this, we get,

$\large{\tt \leadsto \dfrac{2 \pi}{T} = \sqrt{\dfrac{g}{L cos \theta}}}$

Now,Reciprocating it.

$\large{\tt \leadsto \dfrac{T}{2 \pi} = \sqrt{\dfrac{L cos \theta}{g}}}$

$\huge{\boxed{\boxed{T = 2 \pi \sqrt{\dfrac{L cos \theta}{g}}}}}$

Hence derived!.

$\rule{300}{1.5}$

#refer the attachment for figure.

Answer 2

Answer:The time period of conical pendulum is given by :

T = 2π √ l cosO / g

Explanation: