A simple harmonic oscillator has an amplitude A and time period 6π second. Assuming the oscillation starts from its mean position, the time required…

Question

A simple harmonic oscillator has an amplitude A and time period 6π second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=A to x=√32A will be (π)/x s, where x=_______.

MARKS Content Team · Accepted Answer

Correct answer: 2. From phasor diagram particle has to move from P to Q in a circle of radius equal to amplitude of SHM. cosϕ=(√3 A2)/A=√32 ⇒ϕ=(π)/6 Now, (π)/6=ωt ⇒(π)/6=(2π)/Tt ⇒(π)/6=(2π)/(6π)t ⇒t=(π)/2 So, x=2

A simple harmonic oscillator has an amplitude $A$ and time period $6 π$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x = A$ to $x = \frac{\sqrt{3}}{2} A$ will be $\frac{π}{x} s$ , where $x =_______.$

Solution

Relevant Questions

A simple harmonic oscillator has an amplitude A and time period 6π second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=A to x=32A will be πx s, where x=_______.

Solution

Relevant Questions

A simple harmonic oscillator has an amplitude $A$ and time period $6 π$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x = A$ to $x = \frac{\sqrt{3}}{2} A$ will be $\frac{π}{x} s$ , where $x =_______.$