Question
A particle is executing SHM with an amplitude of $4 \mathrm{~cm}$. At the mean position, velocity of the particle is $10 \mathrm{~cm} / \mathrm{s}$. The distance of the particle from the mean position when its speed becomes $5 \mathrm{~cm} / \mathrm{s}$ is(A) $\sqrt{3} \mathrm{~cm}$(B) $\sqrt{5} \mathrm{~cm}$(C) $2 \sqrt{3} \mathrm{~cm}$(D) $2 \sqrt{5} \mathrm{~cm}$
Step 1
Step 1: The displacement in simple harmonic motion is given by $y = A \sin(\omega t + \phi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time, and $\phi$ is the phase constant. Show more…
Show all steps
Your feedback will help us improve your experience
Dheeraj Sharma and 55 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A particle is executing SHM with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is
A particle executes SHM on a line 8 cm long. Its K.E. and P.E. will be equal when its distance from mean position is : A. 4 cm B. 2 cm C. 2√(2) cm D. √(2)cm
A particle executes SHM with an amplitude of $2 \mathrm{~cm} .$ When the particle is at $1 \mathrm{~cm}$ from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is (a) $\frac{1}{2 \pi \sqrt{3}}$ (b) $2 \pi \sqrt{3}$ (c) $\frac{2 \pi}{\sqrt{3}}$ (d) $\frac{\sqrt{3}}{2 \pi}$ (e) $\frac{\sqrt{2}}{\pi}$
Oscillations
Round 2
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD