Question 365: Oscillations

Question

If ' $x$ ', $v$ ' and ' $a$ ' denote the displacement, velocity and acceleration of a particle respectively executing SHM of periodic time $h$ then which one of the following does not change with time?

Accepted Answer

The dimensions of given variables of SHM are as
Displacement, $[x]=\left[M^0 L T^0ight]$
Velocity, $$[\mathrm{v}]=\left[\mathrm{M}^0 \mathrm{LT}^{-1}ight]$$
Acceleration, $[a]=\left[M^0 L T^{-2}ight]$
and time period, $[T]=\left[M^0 L^0 Tight]$
Now, checking each option for these values
For option (a)
$$\frac{[a][T]}{[x]}=\frac{\left[M^0 L T^{-2}ight]\left[M^0 L^0 Tight]}{\left[M^0 L T^0ight]}=\left[M^0 L^0 T^{-1}ight]$$
As it depends on time, so change with it.
For option (b),
$$\begin{aligned}
{[a][T]+2 \pi[v] } & =\left[M^0 L T^{-2}ight]\left[M^0 L^0 Tight]+\left[M^0 L^0 T^{-1}ight] \
& =\left[M^0 L T^{-1}ight]
\end{aligned}$$
It is also dependent on time and hence changes with it.
For option (c),
$$\frac{[a][T]}{[v]}=\frac{\left[M^0 L T^{-2}ight]\left[M^0 L^0 Tight]}{\left[M^0 L T^{-1}ight]}=\left[M^0 L^0 T^0ight]$$
As it is a constant having no dimension, so it does not change with time.
For option (d),
$$\begin{aligned}
	ext { [a] }[t]+4 \pi^2[d]^2 & =\left[M^ho L T^{-2}ight]\left[M^0 L^0 Tight]+\left[M^0 L T^{-1}ight]^2 \
& =\left[L T^{-1}ight]+\left[L^2 T^{-2}ight]
\end{aligned}$$
As, the term is dependent on time, so changes with it.
Also, it is dimensionally incorrect.
Hence, option (c) is correct.

Answer

$\frac{a T}{x}$

Answer

$a t+2 \pi v$

Answer

$\frac{a T}{v}$

Answer

$a T+4 \pi^2 v^2$

Oscillations

If ' x ', v ' and ' a ' denote the displacement, velocity and acceleration of a particle respectively executing SHM of periodic time h then which one of the following does not change with time?

Choose the correct answer: