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Q13.7: NCERT Physics - Part 2 OSCILLATIONS 13.7 Solution: The motion of a particle executing simple harmonic motion is…

⚠️ Common Exam Traps & Conceptual Pitfalls

When you obtain $ an\phi = -1$ from the ratio of the velocity and displacement equations, remember that $ an$ alone does not fix the quadrant; you must also use the signs of $cos\phi$ (from $x(0)=A\cos\phi$) and $sin\phi$ (from $v(0)=-A\omega\sin\phi$) to select the correct angle. Ignoring the negative sign in the velocity expression leads to picking $\phi=3\pi/4$ instead of the correct $\phi=-\pi/4$ (or $7\pi/4$).

The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (_t + _). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is _ cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is � s__. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (_t + _), what are the amplitude and initial phase of the particle with the above initial conditions.

Step-by-Step Verified Solution

**Answer:** Amplitude A=2A = \sqrt{2} cm, initial phase ϕ=3π/4\phi = 3\pi/4. For sine function: B=2B = \sqrt{2} cm, α=π/4\alpha = \pi/4. **Step-by-Step Explanation:** Given: x(0)=1x(0) = 1 cm, v(0)=ωv(0) = \omega cm/s, ω=π\omega = \pi rad/s. For x=Acos⁡(ωt+ϕ)x = A\cos(\omega t + \phi): At t=0t = 0: 1=Acos⁡ϕ1 = A\cos\phi ... (i) Velocity: v=−Aωsin⁡(ωt+ϕ)v = -A\omega\sin(\omega t + \phi). At t=0t = 0: ω=−Aωsin⁡ϕ⇒−1=Asin⁡ϕ\omega = -A\omega\sin\phi \Rightarrow -1 = A\sin\phi ... (ii) Squaring and adding (i) and (ii): A2=1+1=2⇒A=2A^2 = 1 + 1 = 2 \Rightarrow A = \sqrt{2} cm. Dividing (ii) by (i): tan⁡ϕ=−1⇒ϕ=3π/4\tan\phi = -1 \Rightarrow \phi = 3\pi/4 (since sin⁡ϕ<0\sin\phi < 0 and cos⁡ϕ>0\cos\phi > 0) [citation:13]. For x=Bsin⁡(ωt+α)x = B\sin(\omega t + \alpha): At t=0t = 0: 1=Bsin⁡α1 = B\sin\alpha ... (iii) Velocity: v=Bωcos⁡(ωt+α)v = B\omega\cos(\omega t + \alpha). At t=0t = 0: ω=Bωcos⁡α⇒1=Bcos⁡α\omega = B\omega\cos\alpha \Rightarrow 1 = B\cos\alpha ... (iv) Squaring and adding: B2=2⇒B=2B^2 = 2 \Rightarrow B = \sqrt{2} cm. Dividing (iii) by (iv): tan⁡α=1⇒α=π/4\tan\alpha = 1 \Rightarrow \alpha = \pi/4 [citation:13].

All Questions in this Chapter

Q13.1 Q13.2 Q13.3 Q13.4 Q13.5 Q13.6 Q13.7 Q13.8 Q13.9 Q13.10 Q13.11 Q13.12 Q13.13 Q13.14 Q13.15 Q13.16 Q13.17 Q13.18

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