Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The maximum acceleration of a particle in shm is made two times keeping the maximum speed to be constant. this case is possible whena) Amplitude is doubled and frequency is constantb) Amplitude is doubled and frequency is halvedc) Frequency is doubled and amplitude is halvedd) Frequency is doubled and amplitude is constant.

Last updated date: 02nd Aug 2024
Total views: 428.7k
Views today: 9.28k
Verified
428.7k+ views
Hint: Try to write down the maximum acceleration and maximum speed formula in terms of given options, i.e, amplitude and frequency.

Formula Used:
\begin{align} & {{a}_{\max }}={{\omega }^{2}}A \\ & {{v}_{\max }}=\omega A \\ & \omega =2\pi \nu \\ \end{align}

Let the maximum acceleration of the particle ${{a}_{\max }}$, amplitude $A$, frequency $\nu$,maximum speed ${{v}_{\max }}$ angular velocity $\omega$.
initially, \begin{align} & {{a}_{\max }}={{\omega }^{2}}A \\ & {{v}_{max}}=\omega A \\ \end{align}
finally, \begin{align} & {{a}_{\max }}=2{{\omega }^{2}}A \\ & {{v}_{\max }}=\omega A \\ \end{align}
if we convert $\omega$ in terms of frequency, we get $\omega =2\pi \nu$
substituting it in the above formula we get,
\begin{align} & {{a}_{\max }}=4{{\pi }^{2}}{{\nu }^{2}}A \\ & {{v}_{\max }}=2\pi \nu A \\ & \\ \end{align}
Taking option one and four, if only one term is only doubled while the other is constant, we won’t get the required answer. So, answer isn’t option one and four.
Option two, amplitude is doubled, frequency is halved, i.e,
\begin{align} & {{A}_{2}}=2A \\ & {{\nu }_{2}}=\dfrac{\nu }{2} \\ & {{a}_{\max }}=4{{\pi }^{2}}{{(\dfrac{\nu }{2})}^{2}}(2A) \\ & {{a}_{\max }}=2{{\pi }^{2}}{{\nu }^{2}}A \\ \end{align}
Clearly the maximum acceleration is halved but not doubled. so, option two is wrong.
Take option three, frequency is doubled, amplitude is halved, i.e,
\begin{align} & {{\nu }_{2}}=2\nu \\ & A=\dfrac{A}{2} \\ & {{v}_{\max }}=2\pi (2\nu )(\dfrac{A}{2}) \\ & {{v}_{\max }}=2\pi \nu A \\ & {{a}_{\max }}=4{{\pi }^{2}}{{(2\nu )}^{2}}(\dfrac{A}{2}) \\ & {{a}_{\max }}=8{{\pi }^{2}}{{\nu }^{2}}A \\ & \\ \end{align}
Clearly the maximum speed did not change but the maximum acceleration is doubled. so, option c) is right.

Note: Don’t get confused with acceleration and maximum acceleration or speed and maximum speed. The formula of both the terms are different. Also be careful deriving frequency from angular velocity.