1.4 Angular shm  (Page 2/2)

SHM Equation

$$\frac{d^2\theta }{d{t}^2}+{\omega }^2\theta =0$$

Angular velocity

Angular velocity differs to angular frequency (ω). In the case of linear SHM, we had compared linear SHM with uniform circular motion (UCM). It was found that projection of UCM on an axis is equivalent description of linear SHM. It emerged that angular frequency is same as the magnitude of constant angular velocity of the equivalent UCM.

The angular velocity of the body under angular oscillation, however, is different. Importantly, angular velocity of SHM is not constant – whereas angular frequency is constant.

The angular velocity in angular SHM is obtained either as the solution of equation of motion or by differentiating expression of angular displacement with respect to time. Clearly, we need to have a different symbol to represent angular velocity in angular SHM. Let us denote this by the differential expression itself “ $d\theta ∕dt$ ”, which is not equal to “ω”.

$$\frac{d\theta }{dt}=\omega \sqrt{\left({\theta }_0^2-{\theta }^2\right)}=\omega {\theta }_0\cos \left(\omega t+\phi \right)$$

Angular acceleration

$$\alpha =-{\omega }^2\theta =-\frac{k}{I}\theta =-{\omega }^2{\theta }_0\sin \left(\omega t+\phi \right)$$

Torque

$$\tau =I\alpha =-k\theta =-I{\omega }^2\theta ==-I{\omega }^2{\theta }_0\sin \left(\omega t+\phi \right)$$

Frequency, angular frequency, time period

The angular frequency is given by :

$$\omega =\sqrt{\left(\frac{k}{I}\right)}=\sqrt{\left|\frac{\text{angular acceleration}}{\text{angular displacement}}\right|}$$

Time period is obtained from the defining relation :

$$T=\frac{2\pi }{\omega }=2\pi \sqrt{\left(\frac{I}{k}\right)}=2\pi \sqrt{\left|\frac{\text{angular displacement}}{\text{angular acceleration}}\right|}$$

Frequency is obtained from the defining relation :

$$\nu =\frac{1}{T}=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\left(\frac{k}{I}\right)}=\frac{1}{2\pi }\sqrt{\left|\frac{\text{angular acceleration}}{\text{angular displacement}}\right|}$$

Kinetic energy

In terms of angular displacement :

$$K=\frac{1}{2}I{\left(\frac{d\theta }{dt}\right)}^2=\frac{1}{2}I{\omega }^2\left({\theta }_0^2-{\theta }^2\right)$$

In terms of time :

$$K=\frac{1}{2}I{\omega }^2{\theta }_0^2{\cos }^2\left(\omega t+\phi \right)$$

Potential energy

In terms of angular displacement :

$$U=\frac{1}{2}k{\theta }^2=\frac{1}{2}I{\omega }^2{\theta }^2$$

In terms of time :

$$U=\frac{1}{2}I{\omega }^2{\theta }_0^2{\sin }^2\left(\omega t+\phi \right)$$

Mechanical energy

$$E=K+U$$

$$E=\frac{1}{2}I{\omega }^2\left({\theta }_0^2-{\theta }^2\right)+\frac{1}{2}I{\omega }^2{\theta }^2=\frac{1}{2}I{\omega }^2{\theta }_0^2$$

Example

Problem 1: A body executing angular SHM has angular amplitude of 0.4 radian and time period of 0.1 s. If angular displacement of the body is 0.2 radian from the center of oscillation at time t = 0, then write the equation of angular displacement.

Solution : The equation of angular displacement is given by :

$$\theta ={\theta }_0\sin \left(\omega t+\phi \right)$$

Here,

$${\theta }_0=0.4\text{radian}$$

Also, time period is given. Hence, we can determine angular frequency, ω, as :

$$\Rightarrow \omega =\frac{2\pi }{T}=\frac{2\pi }{0.1}=20\pi \text{radian/s}$$

Putting these values, the expression of angular displacement is :

$$\Rightarrow \theta =0.4\sin \left(20\pi t+\phi \right)$$

We, now, need to determine the phase constant for the given initial condition. At t = 0, θ = 0.2 radian. Hence,

$$\Rightarrow 0.2=0.4\sin \left(20\pi X0+\phi \right)=0.4\sin \phi$$

$$\Rightarrow \sin \phi =\frac{0.2}{0.4}=\frac{1}{2}=\sin \frac{\pi }{6}$$

$$\Rightarrow \phi =\frac{\pi }{6}$$

Putting in the expression, the angular displacement is given by :

$$\Rightarrow \theta =0.4\sin \left(20\pi t+\frac{\pi }{6}\right)$$

Moment of inertia and angular shm

Angular SHM provides a very effective technique for measuring moment of inertia. We can have a set up of torsion pendulum for a body, whose MI is to be determined. Measuring time period of oscillation, we have :

$$T=2\pi \sqrt{\left(\frac{I}{k}\right)}$$

Squaring both sides and arranging,

$$\Rightarrow k{T}^2=4{\pi }^{2}I$$

$$\Rightarrow I=\frac{k{T}^2}{4{\pi }^2}$$

Generally, we do not use this equation in the present form as it involves unknown spring constant, “k”. Actually, we carry out experiment for measuring time period first with a regularly shaped object like a rod of known dimensions and mass. This allows us to determine spring constant as MI of the object is known. Then, we determine time period with the object whose MI is to be determined.

Let subscripts “1” and “2” denote objects of known and unknown MIs respectively. Then,

$$\Rightarrow \frac{I_1}{I_2}=\frac{T_1^2}{T_2^2}$$

$$\Rightarrow I_2=\frac{I_1{T}_2^2}{T_1^2}$$

Example

Problem 2: A uniform rod of mass 1 kg and length 0.24 m, hangs from the ceiling with the help of a metallic wire. The time period of SHM is measured to be 2 s. The rod is replaced by a body of unknown shape and mass and the time period for the set up is measured to be 4 s. Find the MI of the body.

Solution : Let subscripts “1” and “2” denote objects of known and unknown MIs. The MI of the rod is calculated, using formula as :

$$\Rightarrow I_1=\frac{m{L}^2}{12}=\frac{1X{0.24}^2}{12}=0.0048kg-m^2$$

The MI of the second body is given by :

$$\Rightarrow I_2=\frac{0.0048X{4}^2}{2^2}=0.0048X4=0.0192kg-m^2$$