OSCILLATIONS
Very Short Question Answers
Q1. Give two examples of periodic motion which are not oscillatory?
Ans.- Revolution of planets around the sun.
- The revolution of the electrons around the nucleus
Q2. The displacement in S.H.M. is given by y = a sin (20t +4). What is the displacement whenit is increased by 2?/? ?
Ans. In S.H.M, the particle comes to same position, for every time period 𝑇 = 2π/ω. So
displacement of the particle remains same even after time is increased by 2π/ω
Q3. A girl is swinging seated in a swing. What is the effect on the frequency of oscillation if the stands?
Ans. When the girl stands up, her centre of mass moves up and length of swing decreases. Therefore, frequency of oscillation increases
Q4. The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hollow sphere?
Ans. When water begins to drain out of sphere, its C.M shifts downward. The length of a pendulum and Time period also increases. When the entire water is drained out of the sphere, its C.M shift to entre of sphere and the time period (T) attains its initial value.
Q5. The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminum?
Ans. Time period remains same because time period does not depend up on the nature of the material of the bob.
Q6. Will a pendulum clock gain or lose time when taken to the top of a mountain?
Ans.
on the top of mountain, value of g is less therefore period of oscillation increases and clock loses time.
Q7. A pendulum clock gives correct time at the equator. Will it gain or lose time if it is taken to the poles? If so, why?
Ans.
on the top of mountain, value of g is less therefore period of oscillation increases and clock loses time.
Q8. What fraction of the total energy is K.E when the displacement is one half of a amplitude of particle executing S.H.M.?
Ans.
Q9. What happens to the energy of a simple harmonic oscillator if its amplitude is doubled?
Ans.
Q10. Can the pendulum clocks be used in an artificial satellite?
Ans. No, in artificial satellite the acceleration due to gravity is zero (𝒈 = 𝒐). So simple pendulum cannot oscillate when it is used in artificial satellite.
Short Question Answers
Essay Question Answers
Q1. Define Simple Harmonic Motion?Show that the projection of uniform circular motion on any diameter is simple harmonic?
Ans. Simple Harmonic Motion: The 'to and fro motion' of a particle about a fixed point is said to be 'simple harmonic motion'. when
- The acceleration is directly proportional to its displacement, in opposite.
- The acceleration is always towards the fixed point
Proof: Consider a particle $P$ moving on the circumference of a circle of radius $A$ with uniform angular velocity $\omega$. As particle $P$ moves on the circumference of the circle, projection $N$ moves to and fro on the diameter $YY'$ about the $0$
Displacement : Form the $\triangle OPN$ $$\begin{aligned}
&&\sin\theta={\frac{ON}{OP}}={\frac{y}{A}} \\
&&\mathbf{y}=\mathbf{A}\sin\theta \\
&&\mathrm{y}=\mathrm{A}\sin\omega t \end{aligned}$$
Velocity:$$\begin{aligned}
&&V={\frac{dy}{dt}} \\
&&V={\frac{d}{dt}}(A\sin\omega t) \\
&&V=\:\mathrm{A}\omega\:\mathrm{Cos}\:\omega t
\end{aligned}$$Acceleration:$$\begin{gathered}
a={\frac{dv}{dt}} \\
a={\frac{d}{dt}}(\mathrm{A}\omega\:\mathrm{Cos}\:\omega t) \\
a=-A\omega^{2}\sin\omega t \\
a=-\omega^{2}y \\
a\propto-y \\
\end{gathered}$$
Hence, the motion of $'N'$ is simple harmonic motion.The projection of uniform circular motion on any diameter is simple harmonic.
Q2. Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is second’s pendulum?
Ans. Derivation:A bob of mass $'m'$ oscillating in aplane about thevertical line through the support. There are only two forces acting on the bob.
i)Tension $T$ due to the string, Vertical force due to gravity $mg$
ii) The force $mg$ can be resolved into two components $mg \:cos\theta$ , $mg \sin\theta$
iII) $mg \cos\theta$ balance with the tension $ (T)$
iv)$mg\sinθ$ provides restoring force.
$$\begin{aligned}
F=-\:mg\sin\theta \\
ma=-\:mg\sin\theta \\
a=-\:g\sin\theta \\
a=-g\theta \\
a=-\:g\left({\frac{x}{l}}\right) \\
a=-\left({\frac{g}{l}}\right)x \\
a\propto-x
\end{aligned}$$Where $g$ and $l$ are constants. Hence, the motion of the simple pendulum is simple harmonic.
Equation for Time Period: from the above equations angular velocity $\omega=\sqrt{\frac{g}{\iota}}$
$$\begin{aligned}&\therefore Time\:period\:T=\frac{2\pi}{\omega}\\&T=2\pi\sqrt{\frac{l}{g}}\\\end{aligned}$$
Seconds Pendulum: A simple pendulum whose time period is two seconds is called seconds pendulum
Q3. Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path?
Ans. Consider a particle $P$ moving on the circumference of a circle of radius $A$ with uniform angular velocity $\omega $ .At a time $t$ The displacement of the particle is $\mathbf{y}=A sin(\omega t)$.Hence, velocity is $V =A\omega cos(\omega t)$.
kinetic energy $K=\frac{1}{2}mv^{2}$ $$\begin{array}{}{K=\frac{1}{2}m(\mathrm{A}\omega\:\mathrm{Cos}\:\omega t)^{2}}\\{K=\frac{1}{2}mA^{2}\omega^{2}cos^{2}\theta } \end{array}$$ Work done for a small displacement $ dy $ is given by $dw=Fdy$
The work done is $W=\frac{1}{2}m\omega^{2}y^{2}$,in simple harmonic motion while moving towards extreme positionsfrom mean position,it will in the form of potential energy.$$\begin{array}{}{\therefore U=\frac{1}{2}m\omega^{2}y^{2}}\\{U=\frac{1}{2}mA^{2}\omega^{2}sin^{2}\theta} \end{array}$$Therefore the total energy of the particle is$$\begin{array}{} {E=K+U}\\{=\frac{1}{2}mA^{2}\omega^{2}cos^{2}\theta+\frac{1}{2}mA^{2}\omega^{2}cos^{2}\theta}\\{E=\frac{1}{2}mA^2\omega^2}\end{array}$$The above equation is independent of time.Hence,The total energy of a particle in simple harmonic motion is constant at any point on it's path.
