The amplitude of a particle executing SHM is $3 \mathrm{~cm}$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is: __________ $\mathrm{cm}$
Explanation:
$ \begin{aligned} & K=1.25 U \\\\ & \Rightarrow K+\frac{K}{1.25}=K_{\max } \\\\ & \Rightarrow \frac{9}{5} K=K_{\max } \\\\ & \Rightarrow \frac{9}{5} \frac{1}{2} m v^{2}=\frac{1}{2} m v_{\max }^{2} \\\\ & \Rightarrow \frac{9}{5}\left[\omega \sqrt{A^{2}-x^{2}}\right]^{2}=\omega^{2} A^{2} \\\\ & \Rightarrow 9\left(A^{2}-x^{2}\right)=5 A^{2} \\\\ & \Rightarrow x^{2}=\frac{4 A^{2}}{9} \\\\ & \Rightarrow x=\frac{2 A}{3} \\\\ & \Rightarrow x=2 \mathrm{~cm} \end{aligned} $
In the figure given below, a block of mass $M=490 \mathrm{~g}$ placed on a frictionless table is connected with two springs having same spring constant $\left(\mathrm{K}=2 \mathrm{~N} \mathrm{~m}^{-1}\right)$. If the block is horizontally displaced through '$\mathrm{X}$' $\mathrm{m}$ then the number of complete oscillations it will make in $14 \pi$ seconds will be _____________.
Explanation:
$=2 \times 2=4 \mathrm{~N} / \mathrm{m}$
$ \mathrm{m}=490 \mathrm{gm} $
$ \begin{aligned} & =0.49 \mathrm{~kg} \end{aligned} $
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Keff}}}=2 \pi \sqrt{\frac{0.49 \mathrm{~kg}}{4}}$
$ =2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10} $
No. of oscillation in the $14 \pi$ is
$ \mathrm{N}=\frac{\text { time }}{\mathrm{T}}=\frac{14 \pi}{7 \pi / 10}=20 $
Explanation:
$4{v^2} = 50 - {x^2}$
or $v = {1 \over 2}\sqrt {50 - {x^2}} $
Comparing the above equation with $v = \omega \sqrt {{A^2} - {x^2}} $
$ \Rightarrow \omega = {1 \over 2}$
& $A = \sqrt {50} $
so ${{2\pi } \over T} = {1 \over 2}$
$ \Rightarrow T = 4\pi \sec $
$ = 4 \times {{22} \over 7}\sec $
$T = {{88} \over 7}\sec $
so $x = 88$
The general displacement of a simple harmonic oscillator is $x = A\sin \omega t$. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $t = {T \over \beta }$. The value of $\beta$ is ______________.
Explanation:
$U = {1 \over 2}m{\omega ^2}{A^2}{\sin ^2}\omega t$
So, ${{dU} \over {dt}} = {{m{\omega ^3}{A^2}} \over 2}\sin 2\omega t$
This value will be maximum when $\sin 2\omega t = 1$
or $2\omega t = {\pi \over 2}$
$2 \times {{2\pi } \over T}t = {\pi \over 2}$
$ \Rightarrow t = {T \over 8}$
So $\beta = 8$
A particle of mass 250 g executes a simple harmonic motion under a periodic force $\mathrm{F}=(-25~x)\mathrm{N}$. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.
Explanation:
$F = - 25x$
$.250{{{d^2}x} \over {d{t^2}}} = - 25x$
${{{d^2}x} \over {d{t^2}}} = - 100x$
$ \Rightarrow \omega = 10$ rad/sec
& $\omega A = {v_{\max }}$
$10\,A = 4$
$ \Rightarrow A = 0.4$ m
$ = 40$ cm
A mass m attached to free end of a spring executes SHM with a period of 1s. If the mass is increased by 3 kg the period of the oscillation increases by one second, the value of mass m is ___________ kg.
Explanation:
Finally
$ 2 \pi \sqrt{\frac{m+3}{k}}=1+1=2 $
Equation $\frac{(1)}{(2)}$ gives
$\sqrt{\frac{m}{m+3}}=\frac{1}{2}$
$\therefore m=1 \mathrm{~kg}$
A block of a mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt x}$ in SI unit. The value of $x$ is ____________.
Explanation:
So $T=2 \pi \sqrt{\frac{m}{K_{n e t}}}$
$ \begin{aligned} & =2 \pi \sqrt{\frac{2}{40}} \\\\ & =\frac{\pi}{\sqrt{5}} \end{aligned} $
$\therefore x=5$
The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is $10 \mathrm{~s}$. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt{x}$ s. The value of $x$ will be ________.
Explanation:
$\mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{F}_{\mathrm{B}}$
$\mathrm{g}^{\prime}=\frac{\mathrm{mg}-\mathrm{F}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{B}}}$
$=\frac{\rho_{\mathrm{B}} \mathrm{Vg}-\rho_{\mathrm{w}} \mathrm{Vg}}{\rho_{\mathrm{B}} \mathrm{V}}$
$=\left(\frac{\rho_{\mathrm{B}}-\rho_{\mathrm{w}}}{\rho_{\mathrm{B}}}\right) \mathrm{g}$
$=\frac{5-1}{5} \times \mathrm{g}$
$=\frac{4}{5} \mathrm{~g}$
We know, $T =2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$
$\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{g}^{\prime}}}=\sqrt{\frac{\mathrm{g}}{5} \mathrm{~g}}=\sqrt{\frac{5}{4}}$
$\mathrm{~T}^{\prime}=\mathrm{T} \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5}$
$\mathrm{~T}^{\prime}=5 \sqrt{5}$
The potential energy of a particle of mass $4 \mathrm{~kg}$ in motion along the x-axis is given by $\mathrm{U}=4(1-\cos 4 x)$ J. The time period of the particle for small oscillation $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right) s$. The value of $\mathrm{K}$ is _________.
Explanation:
$U = 4(1 - \cos 4x)$
$ \Rightarrow F = - {{dU} \over {dx}} = - (4)(4\sin 4x)$
$ = - 16\sin 4x$
as small x
$F = - 16(4x) = - 64x \equiv - kx$
$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{4 \over {64}}} = {\pi \over 2}$
$ \Rightarrow K = 2$
A mass $0.9 \mathrm{~kg}$, attached to a horizontal spring, executes SHM with an amplitude $\mathrm{A}_{1}$. When this mass passes through its mean position, then a smaller mass of $124 \mathrm{~g}$ is placed over it and both masses move together with amplitude $A_{2}$. If the ratio $\frac{A_{1}}{A_{2}}$ is $\frac{\alpha}{\alpha-1}$, then the value of $\alpha$ will be ___________.
Explanation:
$(0.9){A_1}\sqrt {{K \over {0.9}}} = (0.9 + 0.124){A_2}\sqrt {{K \over {0.9 + 0.124}}} $
${{{A_1}} \over {{A_2}}} = \sqrt {{{0.9 + 0.124} \over {0.9}}} $
$ = \sqrt {{{1.024} \over {0.9}}} $
$ = {\alpha \over {\alpha - 1}}$
$\alpha = 16$
As per given figures, two springs of spring constants $k$ and $2 k$ are connected to mass $m$. If the period of oscillation in figure (a) is $3 \mathrm{s}$, then the period of oscillation in figure (b) will be $\sqrt{x}~ s$. The value of $x$ is ___________.
Explanation:
For case (a),
${K_{eq}} = {{2K} \over 3}$
For case (b),
${K_{eq}} = 3K$
$\because$ $T = 2\pi \sqrt {{m \over K}} $
$\therefore$ ${{{T_a}} \over {{T_b}}} = \sqrt {{{{K_b}} \over {{K_a}}}} $
${3 \over {{T_b}}} = \sqrt {{{3K \times 3} \over {2K}}} = {3 \over {\sqrt 2 }}$
${T_b} = \sqrt 2 $
$x = 2$
A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air jet when it is at 5 cm from its mean position. The new amplitude of vibration is $\sqrt{x}$ cm. The value of x is _____________.
Explanation:
$v = \omega \sqrt {{A^2} - {y^2}} $
$ \Rightarrow 3\omega \sqrt {{{10}^2} - {5^2}} = \omega \sqrt {{{(A')}^2} - {5^2}} $
$ \Rightarrow 9 \times 75 = {(A')^2} - 25$
$ \Rightarrow A' = \sqrt {28 \times 25} $ cm
$ \Rightarrow x = 700$
A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60$^\circ$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms$-$1. (if g = 10 m/s2)
Explanation:
${1 \over 2}m{v^2} = mgl(1 - \cos \theta )$
$ \Rightarrow v = \sqrt {2gl(1 - \cos \theta )} $
$ = \sqrt {2 \times 10 \times 2.5 \times {1 \over 2}} $
$ = 5$ m/s
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ___________ s.
Explanation:
A = 8 cm
T = 6 s
$A\cos \left( {{{2\pi t} \over T}} \right) = {A \over 2}$
$ \Rightarrow {{2\pi t} \over T} = {\pi \over 3}$
or $t = {T \over 6} = 1\,s$
Explanation:
KE = PE
$y = {A \over {\sqrt 2 }} = A\sin \omega t$
$t = {T \over 8} = {T \over x}$
x = 8
Explanation:
${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$
${y_2} = 10\left( {{1 \over 2}\sin 3\pi t + {{\sqrt 3 } \over 2}\cos 3\pi t} \right)$
${y_2} = 10\left( {\cos {\pi \over 3}\sin 3\pi t + \sin {\pi \over 3}\cos 3\pi t} \right)$
${y_2} = 10\left( {3\pi t + {\pi \over 3}} \right)$ $\Rightarrow$ Amplitude = 10
So ratio of amplitudes = ${{10} \over {10}}$ = 1
${x_1} = 5\sin \left( {2\pi t + {\pi \over 4}} \right)$ and ${x_2} = 5\sqrt 2 (\sin 2\pi t + \cos 2\pi t)$. The amplitude of second motion is ................ times the amplitude in first motion.
Explanation:
$ = 10\sin \left( {2\pi t + {\pi \over 4}} \right)$
$\therefore$ ${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$
x(t) = A sin($\omega$t + $\phi$)
If the position and velocity of the particle at t = 0 s are 2 cm and 2$\omega$ cm s$-$1 respectively, then its amplitude is $x\sqrt 2 $ cm where the value of x is _________________.
Explanation:
v(t) = A$\omega$ cos ($\omega$t + $\phi$)
2 = A sin$\phi$ ...... (1)
2$\omega$ = A$\omega$ cos$\phi$ ....... (2)
From (1) and (2)
tan$\phi$ = 1
$\phi$ = 45$^\circ$
Putting value of $\phi$ in equation (1),
$2 = A\left\{ {{1 \over {\sqrt 2 }}} \right\}$
$A = 2\sqrt 2 $
$ \therefore $ x = 2
Explanation:
$\mu$ = reduced mass
springs are in series connection
${k_{eq}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$
${k_{eq}} = {{k \times 4k} \over {5k}} = {{4k} \over 5}$
${k_{eq}} = {{4 \times 20} \over 5}$ N/m = 16 N/m
$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = {{0.2 \times 0.8} \over {0.2 + 0.8}} = 0.16$ kg
$\omega = \sqrt {{{16} \over {0.16}}} = \sqrt {100} = 10$
Explanation:
Applying work energy theorem :
wg + wT = $\Delta$K
$-$mgl(1 $-$ cos60$^\circ$) = ${1 \over 2}$mv2 $-$ ${1 \over 2}$mu2
v2 = u2 $-$ 2gl(1 $-$ cos60$^\circ$)
v2 = 9 $-$ 2 $\times$ 10 $\times$ 0.5$\left( {{1 \over 2}} \right)$
v2 = 4
v = 2 m/s
Explanation:
X = A sin ($\omega$t + $\phi$) ($\phi$ = 0 at M.P.)
$ \Rightarrow $ ${A \over 2} = A\sin {{2\pi } \over T}t$
$ \Rightarrow $ ${{2\pi } \over 2}t = {\pi \over 6}$
$ \Rightarrow $ $t = {1 \over 6}$
$ \therefore $ a = 6
Explanation:
${K_{e{q_{series}}}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}} = {k \over 2}$
${T_b} = 2\pi \sqrt {{M \over {k/2}}} = 2\pi \sqrt {{{2M} \over k}} $
${T_b} = \sqrt 2 {T_a}$
${{{T_b}} \over {{T_a}}} = \sqrt 2 $
$ \therefore $ x = 2
Explanation:
$V = \omega \sqrt {{a^2} - {x^2}} $
Given, $V = {{{V_{\max }}} \over 2} = {{A\omega } \over 2}$
${{{A^2}{\omega ^2}} \over 4} = {\omega ^2}{a^2} - {\omega ^2}{x^2}$
$x = {{\sqrt 3 } \over 2}a$
Explanation:
${5 \over 8}$ oscillation = ${1 \over 2}$ oscillation + ${1 \over 8}$ oscillation
From figure, A to B = ${1 \over 2}$ oscillation and B to C is ${1 \over 8}$ oscillation.
$ \therefore $ $\pi { + \,\theta = \omega t} $
$ \Rightarrow $ $\pi { + {{\pi{} } \over 6}} = \omega t$
$ \Rightarrow $ ${{7\pi {} } \over 6} = \left( {{{2\pi {} } \over T}} \right)t$
$ \Rightarrow $ t = ${{7T} \over {12}}$
As shown in the figures, a uniform rod OO' of length l is hinged at the point O and held in place vertically between two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end (O') of the rod, as shown in Fig. 1 and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is f₁. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is f₂. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $ \frac{f_1}{f_2} $ is:
2
$\sqrt{2}$
$\sqrt{\frac{5}{2}}$
$\sqrt{\frac{2}{5}}$
A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency $\omega$ = ($\pi$/3) rad/s. Simultaneously, at t = 0, a small pebble is projected with speed v from point P at an angle of 45$^\circ$ as shown in the figure. Point O is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is (take g = 10 m/s2)
The phase space diagram for a ball thrown vertically up from ground is
The phase space diagram for simple harmonic motion is a circle centred at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total mechanical energies respectively. Then
Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is
If the total energy of the particle is E, it will perform periodic motion only if
For periodic motion of small amplitude A, the time period T of this particle is proportional to
The acceleration of this particle for $|x| > {X_0}$ is
The mass M shown in the figure below oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is
A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is
The $x$-$t$ graph of a particle undergoing simple harmonic motion is shown in the figure. The acceleration of the particle at $t=4/3$ s is
Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 $\times$ 4 matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) | Potential energy of a simple pendulum (y-axis) as a function of displacement (x) axis | (P) |  |
| (B) | Displacement (y-axis) as a function of time (x-axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x-direction | (Q) |  |
| (C) | Range of a projectile (y-axis) as a function of its velocity (x-axis) when projected at a fixed angle | (R) |  |
| (D) | The square of the time period (y-axis) of a simple pendulum as a function of its length (x-axis) | (S) |  |
A small body attached to one end of a vertically hanging spring is performing SHM about its mean position with angular frequency $\omega$ and amplitude $a$. If at a height $y^{\prime}$ from the mean position, the body gets detached from the spring, calculate the value of $y^{\prime}$ so that the height $\mathrm{H}$ attained by the mass is maximum. The body does not interact with the spring during its subsequent motion after detachment $\left(a \omega^{2}>g\right)$
Explanation:
Here, we will use the formula for apparent weight in elevator as the elevator has some acceleration.
The normal force (apparent weight) adjusts for the elevations acceleration
relative to gravity.
Apparent weight: $w=m(g+a)$, where $a$ is the elevator's acceleration.
Here, elevator's position, $y$ is given so we can find elevator's acceleration using
$ a=\frac{d^2 y}{d t^2} $
Given, $m=50 \mathrm{~kg}$
$ \begin{aligned} & \text { and } y=8\left[1+\sin \left(\frac{2 \pi t}{T}\right)\right] \\ & T=40 \pi \mathrm{~s} . \\ & \Rightarrow y=8(1+\sin (\omega t)) \quad\left(as, \omega=\frac{2 \pi}{T}\right) \\ & \omega=\frac{2 \pi}{40 \pi}=\frac{1}{20} \\ & a=\frac{d^2 y}{d t^2}=-8 \omega^2 \sin (\omega t) \\ & \left|a_{\text {max }}\right|=8 \omega^2=8\left(\frac{1}{20}\right)^2=8 \times \frac{1}{400} \\ & \left|a_{\text {max }}\right|=\frac{1}{50} \end{aligned} $
$W_{\text {max }}=m\left(g+a_{\text {max }}\right)$ (when lift goes up)
and. $W_{\text {min }}=m\left(g-a_{\text {max }}\right)$ (when lift goes down)
so
$ \begin{aligned} \Delta w & =w_{\text {max }}-w_{\text {min }}=2 m a_{\text {max }} \\ & =2 \times 50 \times \frac{1}{50} \Rightarrow \Delta w=2 \mathrm{~N} \end{aligned} $
Explanation:
At T t$_0$ = 0
Before collision
After collision
$\begin{aligned} \mathrm{v}_{\mathrm{CM}} & =\frac{\mathrm{m} \cdot \frac{\mathrm{a} \omega}{2}+\mathrm{m} \cdot \mathrm{a} \omega}{\mathrm{m}+\mathrm{m}} \\ \mathrm{v}_{\mathrm{CM}} & =\frac{3 \mathrm{a} \omega}{4} \\ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =\frac{3}{4} \\ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =0.75 \end{aligned}$
Explanation:
$\mathrm{t}_0=\frac{\pi}{2 \omega}=\frac{\mathrm{T}}{4}$
Particles are at extreme position
After collision
in C-frame
using WET,
$\begin{aligned} & \mathrm{W}_{\text {spring }}=\Delta \mathrm{K} \\ & \frac{1}{2} \mathrm{k}(2 \mathrm{~b})^2-\frac{1}{2} \mathrm{k}(2 \mathrm{a})^2=2 \times \frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{a} \omega}{4}\right)^2 \quad(\mathrm{k}=\text { spring constant }) \\ & 4 \mathrm{~kb}^2-4 \mathrm{ka}^2=2 \times \mathrm{m} \times \frac{\mathrm{a}^2}{16} \times \frac{2 \mathrm{k}}{\mathrm{m}} \\ & 4 \mathrm{~b}^2=\frac{17}{4} \mathrm{a}^2 \\ & \frac{4 \mathrm{~b}^2}{\mathrm{a}^2}=4.25 \end{aligned}$
Explanation:
So $a_x=\frac{d^2 x}{d t^2}=-x$
$ \Rightarrow x=A_x \sin \left(\omega t+\phi_x\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s}) $
and $v_x=A_x \omega \cos \left(\omega t+\phi_x\right)$
at $t=0, x=\frac{1}{\sqrt{2}} \mathrm{~m}$ and $v_x=-\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\frac{1}{\sqrt{2}}=A_x \sin \phi_x$
and $-\sqrt{2}=A_x \cos \phi_x$
$ \begin{aligned} & \Rightarrow \tan \phi_x=-\frac{1}{2} ......(1) \\ & \text { and } A_x=\sqrt{\frac{5}{2}} \mathrm{~m} ......(2) \end{aligned} $
Similarly
$ \begin{aligned} & F_y=-y=m a_y . \\\\ & \Rightarrow \frac{d^2 y}{d t^2}=-y \end{aligned} $
So, $y=A_y \sin \left(\omega \mathrm{t}+\phi_{\mathrm{y}}\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s})$ and
$v_y=A_y \omega \cos \left(\omega t+\phi_y\right)$
at $t=0,y=\sqrt{2} \mathrm{~m}$ and $v_y=\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\sqrt{2}=A_y \sin \phi$
and $\sqrt{2}=A_y \cos \phi$
$ \Rightarrow \phi=\frac{\pi}{4} \text { and } A_y=2 $
So,
$\left(x v_y-y v_x\right) $
$ =\sqrt{\frac{5}{2}} \sin \left(\omega t+\phi_x\right) \times 2 \cos \left(\omega t+\phi_y\right)-2 \sin \left(\omega t+\phi_y\right) \times \sqrt{\frac{5}{2}} \cos \left(\omega t+\phi_x\right) $
$ =\sqrt{\frac{5}{2}} \times 2\left(\sin \left(\omega t+\phi_x\right) \cos \left(\omega t+\phi_y\right)-\sin \left(\omega t+\phi_y\right) \times \cos \left(\omega t+\phi_x\right)\right. $
$ =\sqrt{10} \sin \left(\phi_x-\phi_y\right)$
$ =\sqrt{10}\left(\sin \phi_x \cos \phi_y-\cos \phi_x \sin \phi_y\right) $
$ =\sqrt{10}\left(\frac{1}{\sqrt{5}} \times \frac{1}{\sqrt{2}}-\left(-\frac{2}{\sqrt{5}}\right) \times \frac{1}{\sqrt{2}}\right) $
$=3$
On a frictionless horizontal plane, a bob of mass $m=0.1 \mathrm{~kg}$ is attached to a spring with natural length $l_{0}=0.1 \mathrm{~m}$. The spring constant is $k_{1}=0.009 \,\mathrm{Nm}^{-1}$ when the length of the spring $l>l_{0}$ and is $k_{2}=0.016 \,\mathrm{Nm}^{-1}$ when $l < l_{0}$. Initially the bob is released from $l=$ $0.15 \mathrm{~m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) s$, then the integer closest to $n$ is __________.
Explanation:
$ \begin{aligned} & =\mathrm{T}_1+\mathrm{T}_2 \\\\ & =\pi \sqrt{\frac{m}{K_1}}+\pi \sqrt{\frac{m}{K_2}} \\\\ & =\pi \sqrt{\frac{0.1}{0.009}}+\pi \sqrt{\frac{0.1}{0.016}} \\\\ & =\frac{\pi}{0.3}+\frac{\pi}{0.4} \\\\ & =\frac{\pi(0.4+0.3)}{0.12} \\\\ & =\frac{70 \pi}{12} \\\\ & =5.83 \pi \text { seconds } \\\\ & \simeq 6 \pi \text { seconds } \end{aligned} $
Explanation:
When mass m is pulled by a force F, the wire elongation x, length l, cross-sectional area A, and Young's modulus of wire material Y are related by $Y = {{F/A} \over {x/l}}$ i.e.,
$F = (Y\,A/l)x$.
The restoring force by the wire is equal but opposite to F i.e., Fr = $-$F. Apply Newton's second law to get
$m{d^2}x/d{t^2} = - (YA/l)x = - {\omega ^2}x$.
This equation represents SHM with an angular frequency $\omega = \sqrt {YA/(lm)} $. Substitute the values to get $Y = {\omega ^2}lm/A = 4 \times {10^9}$ N/m2.