Oscillations (11Acd12)

Sheet 05 Higher Order Thinking Skills.

1. Two simple harmonic motions are represented by the equations \(y_{1}=0.1Sin\left(100\pi t+\frac{\pi}{3}\right)\) and \(y_{2}=0.1Cos\left(\pi t\right)\). What is the phase difference of the velocity of the particle of 1 with respect to the velocity of the particle 2 ?
   
   Ans: \(\Delta \Phi=(-)\frac{\pi}{6}\).
   
   
2. A particle executes S.H.M between \(y=-r\) and \(y=+r\). The time taken for it to go from \(0\) to \(\frac{r}{2}\) is \(T_{1}\) and to go from \(\frac{r}{2}\) to \(r\) is \(T_{2}\). Then how ate \(T_{1}\) and \(T_{2}\) are related?
   
   Ans: \(\frac{T_{2}}{T_{1}}=2\).
   
   
3. Two S.H.Ms are represented by the following equations \(y_{1}=10Sin\frac{\pi}{4}\left(12t+1\right)\) and \(y_{2}=5\left(Sin3\pi t+\sqrt{3}Cos3\pi t\right)\). Find; the ratio of their amplitudes. What are the time periods of the S.H.M’s?
   
   Ans: \(\frac{r_{1}}{r_{2}}=1\) and \(T_{1}=T_{2}=\frac{2}{3}Sec\).
   
   
4. Springs of spring constants \(k\), \(2k\), \(4k\), \(8k\) …………. are connected in series. A mass \(m\) is attached to the lower end of the last spring and the system is allowed to vibrate. What is the time period of oscillations? Given: \(m=40gm\) and \(k=2.0\frac{N}{cm}\).
   
   Ans: \(T=0.126Sec\).
   
   
5. Two bodies \(M\) and \(N\) of equal masses are suspended from two separate massless spring of spring constants \(k_{1}\) and \(k_{2}\) respectively. If the two bodies oscillates vertically, such that their maximum velocities are equal, then find the ratio of the amplitudes of \(M\) and \(N\).
   
   Ans: \(\frac{r_{1}}{r_{2}}=\sqrt{\frac{k_{2}}{k_{1}}}\).
   
   
6. A particle of mass \(m\) is executing oscillations about the origin on X-axis. Its potential energy is \(V=ky^{3}\), where \(k\) is positive constant. If the amplitude of the oscillations is \(r\), how that the time period of the motion is directly proportional to the amplitude of the motion.
   
   Ans: \(T=2\pi\sqrt{\frac{m}{3krSin(\omega t)}}\).
   
   
7. A mass \(M\) is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes S.H.M of time period \(T\). If the mass is increased by \(m\), the time period becomes \(\frac{5T}{3}\). What is the ratio \(\frac{m}{M}\)?
   
   Ans: \(\frac{m}{M}=\frac{16}{9}\).
   
   
8. A simple pendulum has time period \(T_{1}\). The point of suspension is now moved upwards according to the relation \(y=kt^{2}\) where: \(k=1\frac{m}{s^{2}}\) and \(y\) is vertical displacement. The time period now becomes \(T_{2}\). What is the ratio \(\frac{T_{1}^{2}}{T_{2}^{2}}\)? Given: \(g=10\frac{m}{s^{2}}\).
   
   Ans: \(\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{6}{5}\).
   
   
9. A particle at the end of a spring executes simple harmonic motion with period \(t_{1}\), while the corresponding period for another spring is \(t_{2}\). What is the time period of oscillation when the two springs are connected in series?
   
   Ans: \(T=\sqrt{\left(t_{1}^{2}+t_{2}^{2}\right)}\).
   
   
10. A simple pendulum has a time period \(T\). The pendulum is completely immersed in a non-viscous liquid of density \(\sigma\) whilethe density of the material of the bob os simple pendulum is \(\rho\). What will be the time period of the pendulum immersed in to liquid?
    
    Ans: \(T’=T\sqrt{\frac{\rho}{\rho-\sigma}}\).
    
    
11. A uniform spring whose unstretched length is \(l\) has a force constant \(k\). The spring is cut into two pieces of unstretched length \(l_{1}\) and \(l_{2}\), where \(l_{1}=nl_{2}\) and \(n\) is an integer. What are the corresponding force constants \(k_{1}\) and \(k_{2}\) in terms of \(n\) and \(k\)? What is the ratio of \(k_{1}\) and \(k_{2}\)?
    
    Ans: \(k_{1}=\left(\frac{n+1}{n}\right)\), \(k_{2}=(n+1)k\) and \(\frac{k_{1}}{k_{2}}=\frac{1}{n}\).
    
    
12. A horizontal spring and block system of mass \(M\) executes S.H.M. When the block is passing through its equilibrium position, an object of mass \(m\) is put on it and the two moves together. Find the new amplitude and frequency ?
    
    Ans: \(r’=\sqrt{\frac{M}{M+m}}r\) and \(\nu’=\frac{1}{2\pi}\sqrt{\frac{k}{M+m}}\).
    
    
13. An air tight piston of cross-sectional area \(a\), connected to spring of spring constant \(k\) and unstretched length \(l\) separates two region of a cube. The region enclosed by the piston is evacuated and the opposite region is open to atmosphere. How will you use this setup to determine the atmospheric pressure?
    
    Ans: \(P_{atm}=\frac{k(l’-l)}{a}\).
    
    
14. A particle of mass \(m\) is attached to a spring having spring constant \(k\) and has a natural angular frequency \(\omega_{o}\). An external force \(F(t)\propto Cos(\omega t)\) where; \(\omega\neq\omega_{o}\), is applied to the oscillator. How does the time displacement of the oscillator vary?
    
    Ans: \(y\propto \frac{1}{m(\omega_{o}^{2}-\omega^{2})}\).