Question 344: Oscillations

Question

A particle is performing S.H.M. with maximum velocity '$$v$$'. If the amplitude is tripled and periodic time is doubled then maximum velocity will be

Accepted Answer

To solve this problem, we first need to recall the formula that relates the maximum velocity ($$v_{max}$$) of a particle performing simple harmonic motion (S.H.M.) to its amplitude ($$A$$) and its angular frequency ($$\omega$$). The formula for maximum velocity in S.H.M. is given by:
$$
v_{max} = \omega A
$$
Now the angular frequency ($$\omega$$) is related to the period ($$T$$) of the S.H.M. by the relationship:
$$
\omega = \frac{2\pi}{T}
$$
Thus, we can rewrite the expression for maximum velocity as:
$$
v_{max} = A \cdot \frac{2\pi}{T}
$$
In the initial situation, let's denote the amplitude as $$A$$ and the periodic time as $$T$$. The maximum velocity given is $$v$$, so we can write:
$$
v = A \cdot \frac{2\pi}{T}
$$
According to the question, the amplitude is tripled (making it $$3A$$), and the periodic time is doubled (making it $$2T$$). We need to find the new maximum velocity, let's call it $$v'$$:
$$
v' = (3A) \cdot \frac{2\pi}{(2T)}
$$
Simplifying gives us:
$$
v' = 3A \cdot \frac{\pi}{T}
$$
Now, if we recall the original maximum velocity $$v = A \cdot \frac{2\pi}{T}$$, we can express $$\frac{\pi}{T}$$ in terms of $$v$$ and $$A$$:
$$
\frac{\pi}{T} = \frac{v}{2A}
$$
Substituting this into the expression for $$v'$$, we get:
$$
v' = 3A 	imes \frac{v}{2A} = \frac{3}{2}v = 1.5v
$$
Therefore, the new maximum velocity when the amplitude is tripled and the periodic time is doubled is $$1.5v$$. The correct answer is:
Option A $$1.5 \mathrm{~v}$$

Answer

$$1.5 \mathrm{~v}$$

Answer

$$3 \mathrm{~v}$$

Answer

$$2 \mathrm{~v}$$

Answer

$$\mathrm{v}$$

Oscillations

A particle is performing S.H.M. with maximum velocity '$v$'. If the amplitude is tripled and periodic time is doubled then maximum velocity will be

Choose the correct answer: