Yes — when the same pendulum (same length, same bob) is timed for 10 oscillations again and again, the time taken comes out almost the same every time, so the calculated time period is nearly constant.

Conclusion: The time period of a given simple pendulum is constant at a place — this is exactly why pendulums can be used to build reliable clocks.

Fig 8.7 — A simple pendulum

• Changing the length: a longer pendulum takes more time to complete one oscillation, and a shorter pendulum takes less time — so the time period increases as the length increases.
• Changing the bob’s mass (length fixed): using bobs of different mass on a pendulum of the same length gives almost the same time period every time — so mass has no effect on the time period.

Conclusion: The time period of a simple pendulum depends only on its length, not on the mass of the bob. All pendulums of the same length have the same time period at a given place.

When distances are different, comparing only the time taken isn’t fair — instead we compare the distance covered per unit time, which is called speed.

$$\text{Speed} = \dfrac{\text{Total distance covered}}{\text{Total time taken}}$$

Whichever runner or object covers a greater distance in the same unit of time (i.e., has the higher speed) is considered faster — this lets us compare motion across races of different distances.

This activity uses your own local train timetable, so the exact numbers will differ from student to student. Here is a worked sample showing the method to follow:

Suppose a train departs Station A at 10:00 and reaches the next station, 60 km away, at 10:40.

• Time taken $= 10{:}40 – 10{:}00 = 40\text{ min} = \dfrac{40}{60}\text{ h} = 0.67\text{ h}$
• $$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{60\ \text{km}}{0.67\ \text{h}} \approx 90\ \text{km/h}$$

Once you calculate this for 4–5 different trains (Passenger/Express/Superfast) on your table, the train with the highest speed value (highest distance covered per hour) is the fastest one.

• The runners who kept the same speed throughout that stretch were in uniform linear motion — they covered equal distances in equal intervals of time.
• The runners who kept speeding up or slowing down were in non-uniform linear motion — they covered unequal distances in equal intervals of time.

Both groups moved along the same straight road (linear motion), but differed in whether their speed stayed constant or kept changing.

• Use the speed formula: $$\text{Speed} = \frac{\text{Total distance covered}}{\text{Total time taken}} = \frac{150\ \text{m}}{10\ \text{s}} = 15\ \text{m/s}$$
• Convert m/s to km/h by multiplying by $\dfrac{18}{5}$ (since $1\ \text{m/s} = 3.6\ \text{km/h}$): $$15\ \text{m/s} = 15 \times \frac{18}{5}\ \text{km/h} = 54\ \text{km/h}$$

• Speed of Runner 1: $$\frac{400\ \text{m}}{50\ \text{s}} = 8\ \text{m/s}$$
• Speed of Runner 2: $$\frac{400\ \text{m}}{45\ \text{s}} = 8.89\ \text{m/s}\ \text{(approx.)}$$
• Difference in speed: $$8.89\ \text{m/s} – 8\ \text{m/s} = 0.89\ \text{m/s}$$

• Convert the distance to metres: $$360\ \text{km} = 360 \times 1000\ \text{m} = 360000\ \text{m}$$
• Use $\text{Time taken} = \dfrac{\text{Distance covered}}{\text{Speed}}$: $$\text{Time} = \frac{360000\ \text{m}}{25\ \text{m/s}} = 14400\ \text{s}$$
• Convert seconds to hours: $$14400\ \text{s} = \frac{14400}{3600}\ \text{h} = 4\ \text{h}$$

• (i) Speed in km/h: $$\text{Speed} = \frac{180\ \text{km}}{3\ \text{h}} = 60\ \text{km/h}$$
• (ii) Speed in m/s — multiply by $\dfrac{5}{18}$ (since $1\ \text{km/h} = \dfrac{5}{18}\ \text{m/s}$): $$60\ \text{km/h} = 60 \times \frac{5}{18}\ \text{m/s} = 16.67\ \text{m/s}\ \text{(approx.)}$$
• (iii) Distance in 4 h — use $\text{Distance} = \text{Speed} \times \text{Time}$: $$\text{Distance} = 60\ \text{km/h} \times 4\ \text{h} = 240\ \text{km}$$

• Convert the train’s speed to m/s so both speeds are in the same unit: $$72\ \text{km/h} = 72 \times \frac{5}{18}\ \text{m/s} = 20\ \text{m/s}$$
• Compare: $$\text{Train} = 20\ \text{m/s} \qquad \text{Horse} = 18\ \text{m/s}$$ $$20\ \text{m/s} – 18\ \text{m/s} = 2\ \text{m/s}$$

• Car on a straight highway (no traffic): the car can maintain a constant speed for a long stretch, covering equal distances in equal intervals of time. This is uniform linear motion.
• Car in city traffic: the car has to constantly speed up, slow down, and stop at signals or due to congestion — it covers unequal distances in equal intervals of time. This is non-uniform linear motion.

• Since the object is in uniform motion, it must cover equal distances in every equal 10 s interval. Comparing the known values (0→8, 8→?→24, 32→40, 40→?→56), the distance increases by a constant $8\ \text{m}$ every $10\ \text{s}$: $$\text{Speed} = \frac{8\ \text{m}}{10\ \text{s}} = 0.8\ \text{m/s}$$
• At $t = 20\ \text{s}$: distance $= 8 + 8 = 16\ \text{m}$
• At $t = 60\ \text{s}$: distance $= 40 + 8 = 48\ \text{m}$

• The car covers 60 km, then 70 km, then 50 km in three equal 1-hour intervals — these distances are not equal, so the motion is non-uniform.
• Total distance $= 60 + 70 + 50 = 180\ \text{km}$; Total time $= 3\ \text{h}$
• $$\text{Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}} = \frac{180\ \text{km}}{3\ \text{h}} = 60\ \text{km/h}$$

Non-uniform motion is far more common in daily life, because most moving objects around us rarely keep a perfectly constant speed for long — they speed up, slow down, or stop due to traffic, terrain, obstacles, or simply human control.

• A city bus speeding up between stops and slowing down/halting at each stop and traffic signal.
• A cyclist pedalling faster on a flat road and slower while climbing an incline.
• A person walking, who naturally slows down while crossing a road or turning a corner and speeds up on a clear stretch.

• Distance covered in each successive 10 s interval: $6,\ 4,\ 6,\ 5,\ 8,\ 6,\ 7,\ 3,\ 10,\ 5$ metres — these are not equal, so the object is in non-uniform motion.
• Total distance covered $= 60\ \text{m}$; Total time taken $= 100\ \text{s}$
• $$\text{Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}} = \frac{60\ \text{m}}{100\ \text{s}} = 0.6\ \text{m/s}$$

• Time for the first 500 m at 10 m/s: $$t_1 = \frac{500\ \text{m}}{10\ \text{m/s}} = 50\ \text{s}$$
• Time for the next 500 m at 5 m/s: $$t_2 = \frac{500\ \text{m}}{5\ \text{m/s}} = 100\ \text{s}$$
• Time used so far: $t_1 + t_2 = 50 + 100 = 150\ \text{s}$. Time left for the remaining distance $= 200 – 150 = 50\ \text{s}$
• Remaining distance $= 2000\ \text{m} – 500\ \text{m} – 500\ \text{m} = 1000\ \text{m}$
• Speed needed for the remaining 1000 m: $$\text{Speed} = \frac{1000\ \text{m}}{50\ \text{s}} = 20\ \text{m/s}$$
• Average speed for the entire 2 km journey (total distance ÷ total time): $$\text{Average speed} = \frac{2000\ \text{m}}{200\ \text{s}} = 10\ \text{m/s}$$