4 Marks Question

Question 14 Marks

A spring balance has a scale that reads from 0 to 50 kg . The length of the scale is 20 cm . A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s . What is the weight of the body ?

Answer

Given :
$\begin{aligned} m & =50 kg \\ y & =20-0=20 cm \\ y & =20 \times 10^{-2} m \\ T & =0.6 \text { second }\end{aligned}$
For maximum
$
\begin{aligned}
F & =m g \\
& =50 \times 9.8=490.0 N
\end{aligned}
$
For spring
$
\begin{array}{l}
F=k y \\
k=\frac{F}{y} \\
k=\frac{490}{20 \times 10^{-2}}=24.5 \times 10^2 N / m \\
k=2450 N / m
\end{array}
$
We know, that
$
\begin{aligned}
T & =2 \pi \sqrt{\frac{m^{\prime}}{k}} \\
T^2 & =\frac{4 \pi^2 \times m^{\prime}}{k} \\
m^{\prime} & =\frac{T^2 k}{4 \pi^2}
\end{aligned}
$
$
\begin{aligned}
\text { Putting value } m^{\prime} & =\frac{(0.6)^2 \times 2450}{4 \times 9.87} \\
m^{\prime} & =\frac{0.36 \times 2450}{4 \times 9.86}=\frac{882}{39.44}
\end{aligned}
$
$\therefore$$\quad$Weight of the body
$
\begin{aligned}
w & =m^{\prime} g=22.36 \times 9.8 \\
& =219.1 N
\end{aligned}
$

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Question 24 Marks

The acceleration due to gravity on the surface of moon is $1.7 ms^{-2}$. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? ( $g$ on the surface of earth is $9.8 m s ^{-2}$ ).

Answer

Given :
$
\begin{aligned}
g_{m} & =1.7 m / s^2 \\
g_{e} & =9.8 m / s^2 \\
T_{m} & =? \text { and } T_{e}=3.5 \\
T & =2 \pi \sqrt{\frac{l}{g}}
\end{aligned}
$
Freedom for the earth
$T _{ e }=2 \pi \sqrt{\frac{l}{g_{ e }}}$$\ldots$(1)
For moon
$T _m=2 \pi \sqrt{\frac{l}{g_m}}$$\ldots$(2)
Dividing equation (1) by equation (2)
$
\begin{aligned}
\frac{T_m}{T_e} & =\frac{2 \pi \sqrt{\frac{l}{g_m}}}{2 \pi \sqrt{\frac{l}{g_e}}} \\
\frac{T_m}{T_e} & =\sqrt{\frac{g_e}{g_m}} \\
T_m & =T_e \sqrt{\frac{g_e}{g_m}}
\end{aligned}
$
Put value
$
=3.5 \sqrt{\frac{9.8}{1.7}}
$
$\begin{aligned} & =3.5 \sqrt{5.7647}=3.5 \times 2.4 \\ \because \quad T_m & =8.45\end{aligned}$
Therefore, the time period on the surface of the moon.= 8.45 second

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Question 34 Marks

Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the $x$-projection of the radius vector of the revolving particle $P$, in each case.

Answer

From the figure (a)
OP is in the direction of $\frac{\pi}{2}$ angle of $\frac{\pi}{2}$ since the speed the needle with the $x$-axis thus $\phi=+\frac{\pi}{2}$ radian of clock.
Thus the projectile S.H.M. on the axis of OP at time will given the equation
$
\begin{aligned}
x & =A \cos \left(\frac{2 \pi t}{T}+\phi\right) \\
& =3 \cos \left(\frac{2 \pi t}{2}+\frac{\pi}{2}\right) \\
A & =3 cm, T=2 sec \\
x & =3 \cos \left(\pi t+\frac{\pi}{2}\right)=-3 \sin \pi t \\
x & =-3 \sin \pi t cm
\end{aligned}
$
Figure (b) is
$
T=4 s, A=2 m
$
$t=0$, OP makes an angle of $\pi$ with the positive direction of $x$-axis i.e.
$\begin{aligned} \phi & =+\pi \\ x & = A \cos \left(\frac{2 \pi}{T} t+\phi\right) \\ & =2 \cos \left(\frac{2 \pi t}{T}+\pi\right)=-2 \cos \left(\frac{\pi}{2} t\right)\end{aligned}$
$x=-2 \cos \left(\frac{\pi}{2} t\right) m$

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