Question 304: Oscillations

Question

The displacement of a particle performing S.H.M. is given by $$x=5 \sin (3 t+3)$$, where $$x$$ is in $$\mathrm{cm}$$ and $$t$$ is in second. The maximum acceleration of the particle will be

Accepted Answer

To determine the maximum acceleration of a particle performing Simple Harmonic Motion (S.H.M.), we start with the given displacement function:
$$x = 5 \sin (3t + 3)$$
Here, $ x $ is in centimeters and $ t $ is in seconds. In S.H.M., the acceleration $ a $ is directly proportional to the displacement $ x $ and is given by:
$$a = -\omega^2 x$$
where $ \omega $ is the angular frequency. From the given displacement equation, we can see that the angular frequency $ \omega $ is 3. Thus, the maximum displacement $ x_{	ext{max}} $ is 5 cm because the sine function's amplitude is 5.
The maximum acceleration $ a_{	ext{max}} $ occurs when the sine function equals 1 (i.e., at maximum displacement). Hence, we can calculate the maximum acceleration as follows:
$$a_{	ext{max}} = \omega^2 \cdot x_{	ext{max}}$$
Substituting the values:
$$a_{	ext{max}} = 3^2 \cdot 5$$
$$a_{	ext{max}} = 9 \cdot 5$$
$$a_{	ext{max}} = 45 \, \mathrm{cm} \, \mathrm{s}^{-2}$$
Thus, the maximum acceleration of the particle is:
Option C
: $$45 \, \mathrm{cm} \, \mathrm{s}^{-2}$$

Answer

$$15 \mathrm{~cm} \mathrm{~s}^{-2}$$

Answer

$$30 \mathrm{~cm} \mathrm{~s}^{-2}$$

Answer

$$45 \mathrm{~cm} \mathrm{~s}^{-2}$$

Answer

$$90 \mathrm{~cm} \mathrm{~s}^{-2}$$

Oscillations

The displacement of a particle performing S.H.M. is given by $x=5 \sin (3 t+3), where x is in \mathrm{cm} and t$ is in second. The maximum acceleration of the particle will be

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