Name: Superposition of Two or More SHMs
Uploaded: 2019-07-19
Duration: 9 min 11 s
Description: What happens when two or more simple harmonic motions are superimposed? Watch this video to learn what would be the result of the superposition of SHMs.

Question

A particle having mass $10 g$ oscillates according to the equation $x = (2.0 cm) \sin [(100 s^{- 1}) t + \frac{π}{6}] .$ Find

(a) the amplitude, the time period and the spring constant,

(b) the position, the velocity and the acceleration at $t = 0$ .

Accepted Answer

General equation of an SHM is $x = A \sin (ω t + ϕ)$ , where $A$ is amplitude, $ω$ is angular frequency and $ϕ$ is the phase constant.

Comparing the given equation, $x = (2.0 cm) \sin [(100 rad s^{- 1}) t + \frac{π}{6}]$ with this general equation, we get, $A = 2.0 cm, ω = 100 rad s^{- 1}$ and $ϕ = \frac{π}{6}$ .

So, amplitude is $2.0 cm$ .

Time period $T$ is given by, $T = \frac{2 π}{ω} = \frac{2 π}{100} = 6.28 \times 10^{- 2} s$ .

Angular frequency for a spring mass system is $ω = \sqrt{\frac{k}{m}}$ , where $x = (2.0 cm) \sin [(100 s^{- 1}) t + \frac{π}{6}] .$ is the mass of the block and $k$ is the spring constant.
This gives us $k = m ω^{2}$ .

Spring constant is $\Rightarrow \frac{10}{1000} \times {(100)}^{2} = 100 N m^{- 1}$ .

Position of the particle is given by the equation, so for $t = 0 s$ , $x = (2.0 cm) \sin [(100 rad s^{- 1}) \times 0 + \frac{π}{6}] = 2.0 \sin (\frac{π}{6}) cm = 2.0 \times \frac{1}{2} cm = 1.0 cm .$

Velocity of the particle for any time $t$ can be derived from its general equation as $v = \frac{d x}{d t}$ .

$v = \frac{d x}{d t} = \frac{d ((2.0 cm) \sin [(100 rad s^{- 1}) t + \frac{π}{6}])}{d t} = (2.0 cm) \times 100 s^{- 1} \cos ((100 rad s^{- 1}) t + \frac{π}{6}) = 200 \cos ((100 rad s^{- 1}) t + \frac{π}{6}) cm s^{- 1}$

For $t = 0 s$ , $v = 200 \cos ((100 s^{- 1}) \times 0 + \frac{π}{6}) = 200 \cos (\frac{π}{6}) = 200 \times \frac{\sqrt{3}}{2} = 100 \sqrt{3} cm s^{- 1}$

Similarly, acceleration of the particle can be derived from the equation of velocity as $a = \frac{d v}{d t}$ .

$a = \frac{d v}{d t} = \frac{d (200 \cos ((100 rad s^{- 1}) t + \frac{π}{6}) cm s^{- 1})}{d t} = 200 \times 100 s^{- 1} \times - \sin ((100 rad s^{- 1}) t + \frac{π}{6}) cm s^{- 1} = - 200 \sin ((100 rad s^{- 1}) t + \frac{π}{6}) \times 10^{2} cm s^{- 2} = 200 \sin ((100 rad s^{- 1}) t + \frac{π}{6}) m s^{- 2}$

For $t = 0 s$ , $a = - 200 \sin ((100 rad s^{- 1}) \times 0 + \frac{π}{6}) m s^{- 2} = - 200 \sin (\frac{π}{6}) m s^{- 2} = - 100 m s^{- 2}$

Here, the negative sign indicates that the acceleration is opposite to the velocity.

A particle having mass $10 g$ oscillates according to the equation $x = (2.0 cm) \sin [(100 s^{- 1}) t + \frac{π}{6}] .$ Find

(a) the amplitude, the time period and the spring constant,

(b) the position, the velocity and the acceleration at $t = 0$ .

Important Questions on Simple Harmonic Motion

Learn and Prepare for any exam you want !

A particle having mass 10 g oscillates according to the equation x=(2.0 cm)sin100 s-1t+π6. Find (a) the amplitude, the time period and the spring constant, (b) the position, the velocity and the acceleration at t=0.

Important Questions on Simple Harmonic Motion

Learn and Prepare for any exam you want !

A particle having mass $10 g$ oscillates according to the equation $x = (2.0 cm) \sin [(100 s^{- 1}) t + \frac{π}{6}] .$ Find

(a) the amplitude, the time period and the spring constant,

(b) the position, the velocity and the acceleration at $t = 0$ .