Question

A particle executes SHM with amplitude 0.4 m and time period 24 s. The distance travelled by the particle in 10 s is (initially particle is at its mean position)​

Answer 1

Answer:

Explanation:

Hope it helps

Answer 2

Given: Amplitude is 0.4 m and time period is 24 s.

To find: The distance travelled by the particle in 10 s.

Solution:

• According to the formula of Simple Harmonic Motion (SHM),

        $x = A sin (\omega t)$.

• Here, x is the distance travelled by the particle, A is the ampitude of the particle, ω is the angular velocity of the particle and t is the time in seconds.
• To find the angular velocity, we use the formula,

        $\omega = \frac{2\pi }{T}$, where T is the time period.

• Hence, $x = 0.4 * \sin (\frac{2\pi }{24} * 10)$

                        $= 0.4 * \sin (\frac{5}{6} \pi )$

                        $= 0.4 * \frac{1}{2}$

                        $=0.2 m$

Therefore, the distance travelled by the particle is 0.2m.