Q # 1. Name the two characteristics of simple harmonic motion?
Ans. The characteristics of SHM are
- Acceleration of the body is directly proportional to the displacement and is always directed towards mean position:
\[a \propto - \ x\]
Q # 2.Does frequency depend on the amplitude for harmonic oscillator?
Ans. No, frequency of harmonic oscillator does not depend upon its amplitude. The frequency of the oscillator is describe by the relation:
\[f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\]
The above expression shows that the frequency of harmonic oscillator does not depend upon its amplitude. It only depend on its mass and spring constant.
Q # 3. Can we realize an ideal simple pendulum?
Ans. No, we can't realize an ideal simple pendulum. An ideal simple pendulum should consist of a heavy but small metallic bob suspended from a frictionless rigid support by means of long, weightless and inextensible string. These conditions are impossible to attain in nature. So ideal simple pendulum can't be realized.
Q # 4.What is total distance travelled by an object moving with SHM in a time equal to its period, if its amplitude is A?
Ans. The total distance travelled by ab object moving with SHM in its time period is 4A, where A is amplitude of viberation.
Q # 5. What happens to period of simple pendulum if its length is doubled? What happens if the suspended mass is doubled?
Ans. The time period of the simple pendulum is
\[T = 2\pi\sqrt{\frac{l}{g}}\]
Where \(l\) is length of simple pendulum and \(g\) is gravitational constant.
Case 1: If the Length of Simple Pendulum is Doubled
If the length of simple pendulum is doubled, then the time period will be:
\[T^{'} = 2\pi\sqrt{\frac{2l}{g}} = \sqrt{2}\left( 2\pi\sqrt{\frac{l}{g}} \right) = 1.41\ T\]
So if the length of the simple pendulum is doubled, then its time period increase by 1.41 times of initial time period.
Case 2: If the suspended Mass of Simple Pendulum is Doubled
If the mass of bob of simple pendulum is doubled, then there is no effect on time period, because the period is independent of the mass of simple pendulum.
Q # 6. Does the acceleration of simple harmonic oscillator remains constant during its motion? Is the acceleration ever zero? Explain.
Ans. No, the acceleration does not remain constant. The acceleration of simple harmonic oscillator is given by
\[a \propto - \ x\]
This means that acceleration is proportional to the displacement and is always directed towards mean position.
The acceleration becomes zero at mean position \((x = 0)\) and acceleration becomes maximum at extreme position.
Q # 7. What is meant by phase angle? Does it define angle between maximum displacement and the driving force?
Ans. The angle \(\theta\) which specifies the displacement as well as the direction of motion of the point executing SHM is known as phase.
The phase angle does not define angle between maximum displacement and driving force. It is the angle that the rotating radius makes with the reference direction.
Q # 8. Under what condition does the addition of two simple harmonic motions produce a resultant, which is also simple harmonic?
Ans. Addition of two simple harmonic motion produce a resultant, which is also simple harmonic, if the following conditions are fulfilled:
- Simple harmonic motion should be parallel
- Simple harmonic motion should have same frequency
- Simple harmonic motion should have constant phase difference
Q # 9. Show that in SHM, the accleration is zero when the velocity is greatest and the velocity is zero when the acceleration is greatest?
Ans. The expressions of velocity and acceleration of the body executing SHM are as follow:
\[a = - \omega^{2}x\]
\[v = \ \omega\sqrt{x_{0}^{2} - x^{2}}\]
At Mean Position i.e., x = 0
Acceleration of SHO: \(a = - \omega^{2}(0)\) = 0
Velocity of SHO: \(v = \ \omega\sqrt{x_{0}^{2} - (0)^{2}} = \omega x_{0}\)
So at mean position, the acceleration of SHO is zero but velocity is greatest
At Extreme Position i.e., (\(x = x_{0}\))
Acceleration of SHO: \(a = - \omega^{2}(x_{0}\))
So at extreme position, the velocity of SHO is zero but acceleration is greatest
Q # 10.In relation to SHM, explain the equation:
(i) \(\mathbf{y =}\mathbf{A}\mathbf{\ }\mathbf{\sin}\left(
\mathbf{\omega t + \varphi} \right)\)
(ii) \(\mathbf{a =}\mathbf{-}\mathbf{\
}\mathbf{\omega}^{\mathbf{2}}\mathbf{x}\)
Ans.
- \(y\) is instantaneous displacement,
- \(A\) is amplitude,
- \(\varphi\) is initial phase and
- \(\omega t\) is the angle subtended in time t
- \(a\) is acceleration,
- \(\omega\) is angular frequency and
- \(y\) is instantaneous displacement
Q # 11. Explain the relation between the total energy, Potential energy and kinetic energy for a body oscillating with SHM.
Ans. The total energy \(T{.E}_{\ }\) of the mass spring system at any instant of time is described as the sum of potential energy and kinetic energy at that instant. The instantaneous P.E and K.E of SHO is:
\(P{.E}_{\ } = \frac{1}{2}kx^{2}\)
\(K.E_{\ } = \frac{1}{2}k\left( x_{0}^{2} - x^{2} \right)\)
Thus
\[{T.E}_{\ } = {P.E}_{\ } + {K.E}_{\ }\]
\[{\Longrightarrow T.E}_{\ }\ n = \frac{1}{2}kx^{2} + \frac{1}{2}k\left( x_{0}^{2} - x^{2} \right) = \frac{1}{2}kx^{2} + \frac{1}{2}kx_{0}^{2} - \frac{1}{2}kx^{2}\]
\[{\Longrightarrow T.E}_{\ } = \frac{1}{2}kx_{0}^{2}\ (Constant)\]
Thus total energy of the object executing SHM remains constant.
Q # 12. Describe some common phenomenon in which resonance plays an important role.
Ans. There are some common phenomenon in which the resonance plays an important role such that:
- Cooking of food in microwave oven
- Playing Musical Instruments
Q # 13. In a mass spring system is hung vertically and set into oscillations, why does the motion eventually stop?
Ans. If the mass spring system is hung vertically and set into oscillation, the motion eventually stops due to friction and air resistance and some other damping forces.
.png)
0 Comments